ML Aggarwal Solutions for Class 9 Maths Chapter 3 – Expansions are provided here to help students prepare and excel in their exams. This chapter mainly deals with problems based on expansions. Experts tutors have formulated the solutions in a step by step manner for students to grasp the concepts easily. From the exam point of view, solving the problems on a regular basis, students can improve their conceptual knowledge. The solutions contain brief step wise explanations, which are purely based on the latest syllabus of ICSE board. Students can cross check the answers with the solutions designed by the experts and understand the other ways of solving problems effortlessly. To know more about these concepts, students can access ML Aggarwal Solutions pdf, from the links which are provided below and start practising offline to secure good marks in the board exams.

Chapter 3 – Expansions contain chapter test and the ML Aggarwal Class 9 Solutions present in this page provide solutions to questions related to each topic present in this chapter.

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Exercise 3.1

**By using standard formulae, expand the following (1 to 9):**

**1. (i) (2x + 7y) ^{2}**

**(ii) (1/2 x + 2/3 y) ^{2}**

**Solution:**

(i) (2x + 7y)^{2}

It can be written as

= (2x)^{2}Â + 2 Ã— 2x Ã— 7y + (7y)^{2}

So we get

= 4x^{2}Â + 28xy + 49y^{2}

(ii) (1/2 x + 2/3 y)^{2}

It can be written as

= (1/2 x)^{2} + 2 Ã— Â½x + 2/3y + (2/3 y)^{2}

So we get

= Â¼ x^{2}Â + 2/3 xy + 4/9 y^{2}

**2. (i) (3x + 1/2x) ^{2}**

**(ii) (3x ^{2}y + 5z)^{2}**

**Solution:**

(i) (3x + 1/2x)^{2}

It can be written as

= (3x)^{2}Â + 2 Ã— 3x Ã— 1/2x + (1/2x)^{2}

So we get

= 9x^{2}Â + 3 + 1/4x^{2}

= 9x^{2}Â + 1/4x^{2}Â + 3

(ii) (3x^{2}y + 5z)^{2}

It can be written as

= (3x^{2}y)^{2} + 2 Ã— 3x^{2}y Ã— 5z + (5z)^{2}

So we get

= 9x^{4}y^{2}Â + 30x^{2}yz + 25z^{2}

**3. (i) (3x â€“ 1/2x) ^{2}**

**(ii) (1/2 x â€“ 3/2 y) ^{2}**

**Solution:**

(i) (3x â€“ 1/2x)^{2}

It can be written as

= (3x)^{2}Â â€“ 2 Ã— 3x Ã— 1/2x + (1/2x)^{2}

So we get

= 9x^{2} â€“ 3 + 1/4x^{2}

= 9x^{2}Â + 1/4x^{2}Â â€“ 3

(ii) (1/2 x â€“ 3/2 y)^{2}

It can be written asÂ

= (1/2 x)^{2} + (3/2 y)^{2} â€“ 2 Ã— Â½ x Ã— 3/2 y

So we get

= Â¼ x^{2}Â + 9/4 y^{2}Â â€“ 3/2 xy

= Â¼ x^{2} â€“ 3/2 xy + 9/4 y^{2}

**4. (i) (x + 3) (x + 5)**

**(ii) (x + 3) (x â€“ 5)**

**(iii) (x â€“ 7) (x + 9)**

**(iv) (x â€“ 2y) (x â€“ 3y)**

**Solution:**

(i) (x + 3) (x + 5)

By further calculation

= x^{2} + (3 + 5) x + 3 Ã— 5

So we get

= x^{2} + 8x + 15

(ii) (x + 3) (x â€“ 5)

By further calculation

= x^{2}Â + (3 â€“ 5)x â€“ 3 Ã— 5

So we get

= x^{2} â€“ 2x â€“ 15

(iii) (x â€“ 7) (x + 9)

By further calculation

= x^{2}Â â€“ (7 â€“ 9)x â€“ 7 Ã— 9

So we get

= x^{2} + 2x â€“ 63

(iv) (x â€“ 2y) (x â€“ 3y)

By further calculation

= x^{2}Â â€“ (2y + 3y)x + 2y Ã— 3y

So we get

= x^{2} â€“ 5xy + 6y^{2}

**5. (i) (x â€“ 2y â€“ z) ^{2}**

**(ii) (2x â€“ 3y + 4z) ^{2}**

**Solution:**

(i) (x â€“ 2y â€“ z)^{2}

It can be written as

= [x + (-2y) + (-z)]^{2}

By further calculation

= (x)^{2} + (-2y)^{2} + (-z)^{2} + 2 Ã— x Ã— (-2y) + 2 Ã— (-2y) Ã— (-z) + 2 Ã— (-z) Ã— x

So we get

= x^{2} + 4y^{2} + z^{2} â€“ 4xy + 4yz â€“ 2zx

(ii) (2x â€“ 3y + 4z)^{2}

It can be written as

= [2x + (-3y) + 4z]^{2}

By further calculation

= (2x)^{2} + (-3y)^{2} + (4z)^{2} + 2 Ã— 2x Ã— (-3y) + 2 Ã— (-3y) Ã— 4z + 2 Ã— 4z Ã— 2x

So we get

= 4x^{2}Â + 9y^{2}Â + 16z^{2} -12xy â€“ 24yz + 16zx

**6. (i) (2x + 3/x â€“ 1) ^{2} **

**(ii) (2/3 x â€“ 3/2x â€“ 1) ^{2}**

**Solution:**

(i) (2x + 3/x â€“ 1)^{2}

It can be written as

= [2x + 3/x + (-1)]^{2}

By further calculation

= (2x)^{2}Â + (3/x)^{2}Â + (-1)^{2} + 2 Ã—2x Ã— 3/x + 2 Ã— 3/x Ã— (-1) + 2 Ã— (-1) Ã— 2x

So we get

= 4x^{2}Â + 9/x^{2}Â + 1 + 12 â€“ 6/x â€“ 4x

= 4x^{2}Â + 9/x^{2}Â + 13 â€“ 6/x â€“ 4x

(ii) (2/3 x â€“ 3/2x â€“ 1)^{2}

It can be written as

= [2/3 x â€“ 3/2x â€“ 1]^{2}

By further calculation

= (2/3 x)^{2}Â + (-3/2x)^{2}Â + (-1)^{2}Â + 2 Ã— 2/3 x Ã— (-3/2x) + 2 Ã— (-3/2x) Ã— (-1) + 2 Ã— (-1) Ã— (2/3 x)

So we get

= 4/9 x^{2} + 9/4x^{2}Â + 1 â€“ 2 + 3/x â€“ 4/3 x

= 4/9 x^{2}Â + 9/4x^{2} â€“ 1 â€“ 4/3 x + 3/x

**7. (i) (x + 2) ^{3}**

**(ii) (2a + b) ^{3}**

**Solution:**

(i) (x + 2)^{3}

It can be written as

= x^{3} + 2^{3} + 3 Ã— x Ã— 2 (x + 2)

By further calculation

= x^{3}Â + 8 + 6x (x + 2)

So we get

= x^{3}Â + 8 + 6x^{2} + 12x

= x^{3}Â + 6x^{2}Â + 12x + 8

(ii) (2a + b)^{3}

It can be written as

= (2a)^{3}Â + b^{3}Â + 3 Ã— 2a Ã— b (2a + b)

By further calculation

= 8a^{3}Â + b^{3} + 6ab (2a + b)

So we get

= 8a^{3}Â + b^{3}Â + 12a^{2}b + 6ab^{2}

**8. (i) (3x + 1/x) ^{3}**

**(ii) (2x â€“ 1) ^{3}**

**Solution:**

(i) (3x + 1/x)^{3}

It can be written as

= (3x)^{3} + (1/x)^{3} + 3 Ã— 3x Ã— 1/x (3x + 1/x)

By further calculation

= 27x^{3}Â + 1/x^{3} + 9 (3x + 1/x)

So we get

= 27x^{3} + 1/x^{3}Â + 27x + 9/x

(ii) (2x â€“ 1)^{3}

It can be written as

= (2x)^{3} â€“ 1^{3} â€“ 3 Ã— 2x Ã— 1 (2x â€“ 1)

By further calculation

= 8x^{3}Â â€“ 1 â€“ 6x (2x â€“ 1)

So we get

= 8x^{3} â€“ 1 â€“ 12x^{2} + 6x

= 8x^{3}Â â€“ 12x^{2} + 6x â€“ 1

**9. (i) (5x â€“ 3y) ^{3}**

**(ii) (2x â€“ 1/3y) ^{3}**

**Solution:**

(i) (5x â€“ 3y)^{3}

It can be written as

= (5x)^{3}Â â€“ (3y)^{3} â€“ 3 Ã— 5x Ã— 3y (5x â€“ 3y)

By further calculation

= 125x^{3} â€“ 27y^{3} â€“ 45xy (5x â€“ 3y)

So we get

= 125x^{3} â€“ 27y^{3} â€“ 225x^{2}y + 135xy^{2}

(ii) (2x â€“ 1/3y)^{3}

It can be written as

= (2x)^{3} â€“ (1/3y)^{3} â€“ 3 Ã— 2x Ã— 1/3y (2x â€“ 1/3y)

By further calculation

= 8x^{3} â€“ 1/27y^{3} â€“ 2x/y (2x â€“ 1/3y)

So we get

= 8x^{3} â€“ 1/27y^{3} â€“ 4x^{2}/y + 2x/3y^{2}

**Simplify the following (10 to 19):**

**10. (i) (a + b) ^{2} + (a â€“ b)^{2}**

**(ii) (a + b) ^{2} â€“ (a â€“ b)^{2}**

**Solution:**

(i) (a + b)^{2} + (a â€“ b)^{2}

It can be written as

= (a^{2} + b^{2} + 2ab) + (a^{2} +b^{2}Â â€“ 2ab)

By further calculation

= a^{2} + b^{2} + 2ab + a^{2} + b^{2} â€“ 2ab

So we get

= 2a^{2}Â + 2b^{2}

Taking 2 as common

= 2 (a^{2} + b^{2})

(ii) (a + b)^{2} â€“ (a â€“ b)^{2}

It can be written as

= (a^{2} + b^{2} + 2ab) â€“ (a^{2}Â + b^{2} â€“ 2ab)

By further calculation

= a^{2} + b^{2} + 2ab â€“ a^{2} â€“ b^{2}Â + 2ab

So we get

= 4ab

**11. (i) (a + 1/a) ^{2} + (a â€“ 1/a)^{2}**

**(ii) (a + 1/a) ^{2} â€“ (a â€“ 1/a)^{2}**

**Solution:**

(i) (a + 1/a)^{2} + (a â€“ 1/a)^{2}

It can be written as

= [a^{2}Â + (1/a)^{2} + 2 Ã— a Ã— 1/a] + [a^{2} + (1/a)^{2} â€“ 2 Ã— a Ã— 1/a]

By further calculation

= [a^{2} + 1/a^{2} + 2] + [a^{2} + 1/a^{2}Â â€“ 2]

So we get

= a^{2} + 1/a^{2} + 2 + a^{2} + 1/a^{2}Â â€“ 2

= 2a^{2}Â + 2/a^{2}

Taking 2 as common

= 2 (a^{2} + 1/a^{2})

(ii) (a + 1/a)^{2} â€“ (a â€“ 1/a)^{2}

It can be written as

= [a^{2}Â + (1/a)^{2} + 2 Ã— a Ã— 1/a] – [a^{2} + (1/a)^{2} â€“ 2 Ã— a Ã— 1/a]

By further calculation

= [a^{2} + 1/a^{2} + 2] – [a^{2} + 1/a^{2}Â â€“ 2]

So we get

= a^{2} + 1/a^{2} + 2 â€“ a^{2} â€“ 1/a^{2} + 2

= 4

**12. (i) (3x â€“ 1) ^{2} â€“ (3x â€“ 2) (3x + 1)**

**(ii) (4x + 3y) ^{2} â€“ (4x â€“ 3y)^{2} â€“ 48xy**

**Solution:**

(i) (3x â€“ 1)^{2} â€“ (3x â€“ 2) (3x + 1)

It can be written as

= [(3x)^{2} + 1^{2} â€“ 2 Ã— 3x Ã— 1] â€“ [(3x)^{2} â€“ (2 â€“ 1) (3x) â€“ 2 Ã— 1]

By further calculation

= [9x^{2} + 1 â€“ 6x] â€“ [9x^{2} â€“ 3x â€“ 2]

So we get

= 9x^{2} + 1 â€“ 6x â€“ 9x^{2} + 3x + 2

= -3x + 3

= 3 â€“ 3x

(ii) (4x + 3y)^{2} â€“ (4x â€“ 3y)^{2} â€“ 48xy

It can be written as

= [(4x)^{2} + (3y)^{2} + 2 Ã— 4x Ã— 4y] â€“ [(4x)^{2} + (3y)^{2} â€“ 2 Ã— 4x Ã— 3y] â€“ 48xy

By further calculation

= [16x^{2}Â + 9y^{2}Â + 24xy] â€“ [16x^{2} + 9y^{2} â€“ 24xy] â€“ 48xy

So we get

= 16x^{2}Â + 9y^{2} + 24xy â€“ 16x^{2}Â â€“ 9y^{2}Â + 24xy â€“ 48xy

= 0

**13. (i) (7p + 9q) (7p â€“ 9q)**

**(ii) (2x â€“ 3/x) (2x + 3/x)**

**Solution:**

(i) (7p + 9q) (7p â€“ 9q)

It can be written as

= (7p)^{2} â€“ (9q)^{2}

= 49p^{2}Â â€“ 81q^{2}

(ii) (2x â€“ 3/x) (2x + 3/x)

It can be written as

= (2x)^{2}Â â€“ (3/x)^{2}

= 4x^{2} â€“ 9/x^{2}

**14. (i) (2x â€“ y + 3) (2x â€“ y â€“ 3)**

**(ii) (3x + y â€“ 5) (3x â€“ y â€“ 5)**

**Solution:**

(i) (2x â€“ y + 3) (2x â€“ y â€“ 3)

It can be written as

= [(2x â€“ y) + 3] [(2x â€“ y) â€“ 3]

= (2x â€“ y)^{2} – 3^{2}

By further calculation

= (2x)^{2} +y^{2} â€“ 2 Ã— 2x Ã— y â€“ 9

So we get

= 4x^{2} + y^{2} â€“ 4xy â€“ 9

(ii) (3x + y â€“ 5) (3x â€“ y â€“ 5)

It can be written as

= [(3x â€“ 5) + y] [(3x â€“ 5) â€“ y]

= (3x â€“ 5^{2}) â€“ y^{2}

By further calculation

= (3x)^{2} + 5^{2} â€“ 2 Ã— 3x Ã— 5 â€“ y^{2}

So we get

= 9x^{2} + 25 â€“ 30x â€“ y^{2}

= 9x^{2} â€“ y^{2} â€“ 30x + 25

**15. (i) (x + 2/x â€“ 3) (x â€“ 2/x â€“ 3)**

**(ii) (5 â€“ 2x) (5 + 2x) (25 + 4x ^{2})**

**Solution:**

(i) (x + 2/x â€“ 3) (x â€“ 2/x â€“ 3)

It can be written as

= [(x â€“ 3) + (2/x)] [(x â€“ 3) â€“ (2/x)]

= (x â€“ 3)^{2} â€“ (2/x)^{2}

Expanding using formula

= x^{2} + 9 â€“ 2 Ã— x Ã— 3 â€“ 4/x^{2}

By further calculation

= x^{2} + 9 â€“ 6x â€“ 4/x^{2}

So we get

= x^{2} â€“ 4/x^{2} â€“ 6x + 9

(ii) (5 â€“ 2x) (5 + 2x) (25 + 4x^{2})

It can be written as

= [5^{2}Â â€“ (2x)^{2}] (25 + 4x^{2})

By further calculation

= (25 â€“ 4x^{2}) (25 + 4x^{2})

So we get

= 25^{2} â€“ (4x^{2})^{2}

= 625 â€“ 16x^{4}

**16. (i) (x + 2y + 3) (x + 2y + 7)**

**(ii) (2x + y + 5) (2x + y â€“ 9)**

**(iii) (x â€“ 2y â€“ 5) (x â€“ 2y + 3)**

**(iv) (3x â€“ 4y â€“ 2) (3x â€“ 4y â€“ 6)**

**Solution:**

(i) (x + 2y + 3) (x + 2y + 7)

Consider x + 2y = a

(a + 3) (a + 7) = a^{2} + (3 + 7) a + 3 Ã— 7

By further calculation

= a^{2} + 10a + 21

Substituting the value of a

= (x + 2y)^{2} + 10 (x + 2y) + 21

By expanding using formula

= x^{2}Â + 4y^{2} + 2 Ã— x Ã— 2y + 10x + 20y + 21

So we get

= x^{2} + 4y^{2} + 4xy + 10x + 20y + 21

(ii) (2x + y + 5) (2x + y â€“ 9)

Consider 2x + y = a

(a + 5) (a â€“ 9) = a^{2} + (5 â€“ 9) a + 5 Ã— (-9)

By further calculation

= a^{2} â€“ 4a â€“ 45

Substituting the value of a

= (2x + y)^{2}Â â€“ 4 (2x + y) â€“ 45

By expanding using formula

= 4x^{2} + y^{2}Â + 2 Ã— 2x Ã— y – 8x â€“ 4y â€“ 45

So we get

= 4x^{2} + y^{2}Â + 4xy â€“ 8x â€“ 4y â€“ 45

(iii) (x â€“ 2y â€“ 5) (x â€“ 2y + 3)

Consider x â€“ 2y = a

(a â€“ 5) (a + 3) = a^{2} + (- 5 + 3) a + (-5) (3)

By further calculation

= a^{2} â€“ 2a â€“ 15

Substituting the value of a

= (x â€“ 2y)^{2} â€“ 2 (x â€“ 2y) â€“ 15

By expanding using formula

= x^{2}Â + 4y^{2}Â â€“ 2 Ã— x Ã— 2y â€“ 2x + 4y â€“ 15

So we get

= x^{2} + 4y^{2}Â â€“ 4xy â€“ 2x + 4y â€“ 15

(iv) (3x â€“ 4y â€“ 2) (3x â€“ 4y â€“ 6)

Consider 3x â€“ 4y = a

(a â€“ 2) (a â€“ 6) = a^{2}Â (- 2 â€“ 6)a + (-2) (-6)

By further calculation

= a^{2}Â â€“ 8a + 12

Substituting the value of a

= (3x â€“ 4y)^{2} â€“ 8 (3x â€“ 4y) + 12

Expanding using formula

= 9x^{2}Â + 16y^{2} â€“ 2 Ã— 3x Ã— 4y â€“ 24x + 32y + 12

So we get

= 9x^{2} + 16y^{2}Â â€“ 24xy â€“ 24x + 32y + 12

**17. (i) (2p + 3q) (4p ^{2} â€“ 6pq + 9q^{2})**

**(ii) (x + 1/x) (x ^{2} – 1 + 1/x^{2})**

**Solution:**

(i) (2p + 3q) (4p^{2} â€“ 6pq + 9q^{2})

It can be written as

= (2p + 3q) [(2p)^{2}Â – 2p Ã— 3q + (3q)^{2}]

By further simplification

= (2p)^{3} + (3q)^{3}

= 8p^{3}Â + 27q^{3}

(ii) (x + 1/x) (x^{2} – 1 + 1/x^{2})

It can be written as

= (x + 1/x) [x^{2} â€“ x Ã— 1/x + (1/x)^{2}]

By further simplification

= x^{3} + (1/x)^{3}

= x^{3}Â + 1/x^{3}

**18. (i) (3p â€“ 4q) (9p ^{2} + 12pq + 16q^{2})**

**(ii) (x â€“ 3/x) (x ^{2}Â + 3 + 9/x^{2})**

**Solution:**

(i) (3p â€“ 4q) (9p^{2} + 12pq + 16q^{2})

It can be written as

= (3p â€“ 4q) [(3p)^{2} + 3p Ã— 4q + (4q)^{2}]

By further simplification

= (3p)^{3} â€“ (4q)^{3}

= 27p^{3} â€“ 64q^{3}

(ii) (x â€“ 3/x) (x^{2}Â + 3 + 9/x^{2})

It can be written as

= (x â€“ 3/x) [x^{2} + x Ã— 3/x + (3/x)^{2}]

By further simplification

= x^{3} â€“ (3/x)^{3}

= x^{3} â€“ 27/x^{3}

**19. (2x + 3y + 4z) (4x ^{2} + 9y^{2} + 16z^{2} â€“ 6xy â€“ 12yz â€“ 8zx).**

**Solution:**

(2x + 3y + 4z) (4x^{2} + 9y^{2} + 16z^{2} â€“ 6xy â€“ 12yz â€“ 8zx)

It can be written as

= (2x + 3y + 4z) ((2x)^{2} + (3y)^{2} + (4z)^{2} â€“ 2x Ã— 3y â€“ 3y Ã— 4z â€“ 4z Ã— 2x)

By further calculation

= (2x)^{3} + (3y)^{3} + (4z)^{3} â€“ 3 Ã— 2x Ã— 3y Ã— 4z

So we get

= 8x^{3} + 27y^{3} + 64z^{3} â€“ 72xyz

**20. Find the product of the following:**

**(i) (x + 1) (x + 2) (x + 3)**

**(ii) (x â€“ 2) (x â€“ 3) (x + 4)**

**Solution:**

(i) (x + 1) (x + 2) (x + 3)

It can be written as

= x^{3} + (1 + 2 + 3)x^{2} + (1 Ã— 2 + 2 Ã— 3 + 3 Ã— 1) x + 1 Ã— 2 Ã— 3

By further calculation

= x^{3} + 6x^{2} + (2 + 6 + 3)x + 6

So we get

= x^{3} + 6x^{2} + 11x + 6

(ii) (x â€“ 2) (x â€“ 3) (x + 4)

It can be written as

= x^{3} + (- 2 â€“ 3 + 4) x^{2} + [(-2) Ã— (-3) + (-3) Ã— 4 + 4 Ã— (-2)]x + (-2) (-3) (4)

By further calculation

= x^{3} â€“ x^{2} + (6 â€“ 12 â€“ 8)x + 24

= x^{3}Â â€“ x^{2} â€“ 14x + 24

**21. Find the coefficient of x ^{2} and x in the product of (x â€“ 3) (x + 7) (x â€“ 4).**

**Solution:**

It is given that

(x â€“ 3) (x + 7) (x â€“ 4)

By further calculation

= x^{3} + (- 3 + 7 â€“ 4) x^{2} + [(-3) (7) + 7 Ã— (-4) + (-4) (-3) + (-3) (7) (-4)]

It can be written as

= x^{3} + 0x^{2}Â + (- 21 â€“ 28 + 12) x + 84

So we get

= x^{3} + 0x^{2} â€“ 37x + 84

Hence, coefficient of x^{2}Â is zero and coefficient of x is â€“ 3.

**22. If a ^{2} + 4a + x = (a + 2)^{2}, find the value of x.**

**Solution:**

It is given that

a^{2} + 4a + x = (a + 2)^{2}

By expanding using formula

a^{2}Â + 4a + x = a^{2} + 2^{2} + 2 Ã— a Ã— 2

By further calculation

a^{2} + 4a + x = a^{2} + 4 + 4a

So we get

x = a^{2} + 4 + 4a â€“ a^{2} â€“ 4a

x = 4

**23. Use (a + b) ^{2} = a^{2} + 2ab + b^{2} to evaluate the following:**

**(i) (101) ^{2}**

**(ii) (1003) ^{2} **

**(iii) (10.2) ^{2}**

**Solution:**

(i) (101)^{2}

It can be written as

= (100 + 1)^{2}

Expanding using formula

= 100^{2} + 1^{2} + 2 Ã— 100 Ã— 1

By further calculation

= 10000 + 1 + 200

= 10201

(ii) (1003)^{2}

It can be written as

= (1000 + 3)^{2}

Expanding using formula

= 1000^{2} + 3^{2} + 2 Ã— 1000 Ã— 3

By further calculation

= 1000000 + 9 + 6000

= 1006009

(iii) (10.2)^{2}

It can be written as

= (10 + 0.2)^{2}

Expanding using formula

= 10^{2} + 0.2^{2} + 2 Ã— 10 Ã— 0.2

By further calculation

= 100 + 0.04 + 4

= 104.04

**24. Use (a â€“ b) ^{2} = a^{2} â€“ 2ab â€“ b^{2} to evaluate the following:**

**(i) (99) ^{2}**

**(ii) (997) ^{2}**

**(iii) (9.8) ^{2}**

**Solution:**

(i) (99)^{2}

It can be written as

= (100 â€“ 1)^{2}

Expanding using formula

= 100^{2} â€“ 2 Ã— 100 Ã— 1 + 1^{2}

By further calculation

= 10000 â€“ 200 + 1

= 9801

(ii) (997)^{2}

It can be written as

= (1000 â€“ 3)^{2}

Expanding using formula

= 1000^{2} â€“ 2 Ã— 1000 Ã— 3 + 3^{2}

By further calculation

= 1000000 â€“ 6000 + 9

= 994009

(iii) (9.8)^{2}

It can be written as

= (10 â€“ 0.2)^{2}

Expanding using formula

= 10^{2} â€“ 2 Ã— 10 Ã— 0.2 + 0.2^{2}

By further calculation

= 100 â€“ 4 + 0.04

= 96.04

**25. By using suitable identities, evaluate the following:**

**(i) (103) ^{3}**

**(ii) (99) ^{3}**

**(iii) (10.1) ^{3}**

**Solution:**

(i) (103)^{3}

It can be written as

= (100 + 3)^{3}

Expanding using formula

= 100^{3} + 3^{3} + 3 Ã— 100 Ã— 3 (100 + 3)

By further calculation

= 1000000 + 27 + 900 Ã— 103

So we get

= 1000000 + 27 + 92700

= 1092727

(ii) (99)^{3}

It can be written as

= (100 â€“ 1)^{3}

Expanding using formula

= 100^{3} â€“ 1^{3} â€“ 3 Ã— 100 Ã— 1 (100 â€“ 1)

By further calculation

= 1000000 â€“ 1 â€“ 300 Ã— 99

So we get

= 1000000 â€“ 1 â€“ 29700

= 1000000 â€“ 29701

= 970299

(iii) (10.1)^{3}

It can be written as

= (10 + 0.1)^{3}

Expanding using formula

= 10^{3}Â + 0.1^{3} + 3 Ã— 10 Ã— 0.1 (10 + 0.1)

By further calculation

= 1000 + 0.001 + 3 Ã— 10.1

So we get

= 1000 + 0.001 + 30.3

= 1030.301

**26. If 2a â€“ b + c = 0, prove that 4a ^{2} â€“ b^{2} + c^{2} + 4ac = 0.**

**Solution:**

It is given that

2a â€“ b + c = 0

2a + c = b

By squaring on both sides

(2a + c)^{2 }= b^{2}

Expanding using formula

(2a)^{2} + 2 Ã— 2a Ã— c + c^{2}Â = b^{2}

By further calculation

4a^{2} + 4ac + c^{2} = b^{2}

So we get

4a^{2} â€“ b^{2}Â + c^{2}Â + 4ac = 0

Hence, it is proved.

**27. If a + b + 2c = 0, prove that a ^{3} + b^{3} + 8c^{3} = 6abc.**

**Solution:**

It is given that

a + b + 2c = 0

We can write it as

a + b = – 2c

By cubing on both sides

(a + b)^{3} = (-2c)^{3}

Expanding using formula

a^{3}Â + b^{3} + 3ab (a + b) = -8c^{3}

Substituting the value of a + b

a^{3}Â + b^{3} + 3ab (-2c) = -8c^{3}

So we get

a^{3} + b^{3}Â + 8c^{3} = 6abc

Hence, it is proved.

**28. If a + b + c = 0, then find the value of a ^{2}/bc + b^{2}/ca + c^{2}/ab.**

**Solution:**

It is given that

a + b + c = 0

We can write it as

a^{3}Â + b^{3} + c^{3} â€“ 3abc = 0

a^{3} + b^{3}Â + c^{3} = 3abc

Now dividing by abc on both sides

a^{3}/abc + b^{3}/abc + c^{3}/abc = 3

By further calculation

a^{2}/bc + b^{2}/ac + c^{2}/ab = 3

Therefore, the value of a^{2}/bc + b^{2}/ca + c^{2}/ab is 3.

**29. If x + y = 4, then find the value of x ^{3} + y^{3} + 12xy â€“ 64.**

**Solution:**

It is given that

x + y = 4

By cubing on both sides

(x + y)^{3} = 4^{3}

Expanding using formula

x^{3} + y^{3} + 3xy (x + y) = 64

Substituting the value of x + y

x^{3} + y^{3} + 3xy (4) = 64

So we get

x^{3}Â + y^{3} + 12xy â€“ 64 = 0

Hence, the value of x^{3}Â + y^{3} + 12xy â€“ 64 is 0.

**30. Without actually calculating the cubes, find the values of:**

**(i) (27) ^{3} + (-17)^{3} + (-10)^{3}**

**(ii) (-28) ^{3} + (15)^{3} + (13)^{3}**

**Solution:**

(i) (27)^{3} + (-17)^{3} + (-10)^{3}

Consider a = 27, b = – 17 and c = – 10

We know that

a + b + c = 27 â€“ 17 â€“ 10 = 0

So a + b + c = 0

a^{3} + b^{3} + c^{3} = 3abc

Substituting the values

27^{3} + (-17)^{3} + (-10)^{3} = 3 (27) (-17) (- 10)

= 13770

(ii) (-28)^{3} + (15)^{3} + (13)^{3}

Consider a = – 28, b = 15 and c = 13

We know that

a + b + c = – 28 + 15 + 13 = 0

So a + b + c = 0

a^{3} + b^{3} + c^{3} = 3abc

Substituting the values

(-28)^{3} + (15)^{3Â }+ (13)^{3} = 3 (- 28) (15) (13)

= – 16380

**31. Using suitable identity, find the value of:**

**Solution:**

Consider x = 86 and y = 14

= x + y

Substituting the values

= 86 + 14

= 100

### Exercise 3.2

**1. If x â€“ y = 8 and xy = 5, find x ^{2} + y^{2}.**

**Solution:**

We know that

(x â€“ y)^{2} = x^{2} + y^{2} â€“ 2xy

It can be written as

x^{2} + y^{2} = (x â€“ y)^{2} + 2xy

It is given that

x â€“ y = 8 and xy = 5

Substituting the values

x^{2} + y^{2} = 8^{2} + 2 Ã— 5

So we get

= 64 + 10

= 74

**2. If x + y = 10 and xy = 21, find 2 (x ^{2} + y^{2}).**

**Solution:**

We know that

(x + y)^{2}Â = x^{2} + y^{2}Â + 2xy

It can be written as

x^{2} + y^{2} = (x + y)^{2} â€“ 2xy

It is given that

(x + y) = 10 and xy = 21

Substituting the values

x^{2} + y^{2} = 10^{2} â€“ 2 Ã— 21

By further calculation

= 100 â€“ 42

= 58

Here

2 (x^{2} + y^{2}) = 2 Ã— 58 = 116

**3. If 2a + 3b = 7 and ab = 2, find 4a ^{2} + 9b^{2}.**

**Solution:**

We know that

(2a + 3b)^{2} = 4a^{2} + 9b^{2} + 12ab

It can be written as

4a^{2} + 9b^{2} = (2a + 3b)^{2} â€“ 12ab

It is given that

2a + 3b = 7

ab = 2

Substituting the values

4a^{2} + 9b^{2} = 7^{2} â€“ 12 Ã— 2

By further calculation

= 49 â€“ 24

= 25

**4. If 3x â€“ 4y = 16 and xy = 4, find the value of 9x ^{2} + 16y^{2}.**

**Solution:**

We know that

(3x â€“ 4y)^{2} = 9x^{2} + 16y^{2} â€“ 24xy

It can be written as

9x^{2} + 16y^{2} = (3x â€“ 4y)^{2} + 24xy

It is given that

3x â€“ 4y = 16 and xy = 4

Substituting the values

9x^{2} + 16y^{2}_{ }= 16^{2} + 24 Ã— 4

By further calculation

= 256 + 96

= 352

**5. If x + y = 8 and x â€“ y = 2, find the value of 2x ^{2} + 2y^{2}.**

**Solution:**

We know that

2 (x^{2} + y^{2}) = (x + y)^{2}Â + (x â€“ y)^{2}

It is given that

x + y = 8 and x â€“ y = 2

Substituting the values

2x^{2}Â + 2y^{2} = 8^{2}Â + 2^{2}

By further calculation

= 64 + 4

= 68

**6. If a ^{2}Â + b^{2} = 13 and ab = 6, find **

**(i) a + b**

**(ii) a â€“ b**

**Solution:**

(i) We know that

(a + b)^{2} = a^{2} + b^{2} + 2ab

Substituting the values

= 13 + 2 Ã— 6

So we get

= 13 + 12

= 25

Here

a + b = Â± âˆš25 = Â± 5

(ii) We know that

(a â€“ b)^{2} = a^{2} + b^{2} â€“ 2ab

Substituting the values

= 13 â€“ 2 Ã— 6

So we get

= 13 â€“ 12

= 1

Here

a â€“ b = Â± âˆš1 = Â± 1

**7. If a + b = 4 and ab = -12, find**

**(i) a â€“ b**

**(ii) a ^{2}Â â€“ b^{2}.**

**Solution:**

(i) We know that

(a â€“ b)^{2} = a^{2} + b^{2} â€“ 2ab

It can be written as

(a â€“ b)^{2} = a^{2}Â + b^{2}Â + 2ab â€“ 4ab

(a â€“ b)^{2} = (a + b)^{2} â€“ 4ab

It is given that

a + b = 4 and ab = – 12

Substituting the values

(a â€“ b)^{2} = 4^{2}Â â€“ 4 (-12)

By further calculation

(a â€“ b)^{2}Â = 16 + 48 = 64

So we get

(a â€“ b) = Â± âˆš64 = Â± 8

(ii) We know that

a^{2}Â â€“ b^{2} = (a + b) (a â€“ b)

Substituting the values

a^{2}Â â€“ b^{2} = 4 Ã— Â±8

a^{2}Â â€“ b^{2} = Â± 32

**8. If p â€“ q = 9 and pq = 36, evaluate**

**(i) p + q**

**(ii) p ^{2} â€“ q^{2}.**

**Solution:**

(i) We know that

(p + q)^{2} = p^{2} + q^{2} + 2pq

It can be written as

(p + q)^{2} = p^{2} + q^{2} â€“ 2pq + 4pq

(p + q)^{2} = (p â€“ q)^{2} + 4pq

It is given that

p â€“ q = 9 and pq = 36

Substituting the values

(p + q)^{2} = 9^{2}Â + 4 Ã— 36

By further calculation

(p + q)^{2} = 81 + 144 = 225

So we get

p + q = Â± âˆš225 = Â± 15

(ii) We know that

p^{2} â€“ q^{2} = (p â€“ q) (p + q)

Substituting the values

p^{2} â€“ q^{2} = 9 Ã—Â±15

p^{2} â€“ q^{2} = Â± 135

**9. If x + y = 6 and x â€“ y = 4, find **

**(i) x ^{2} + y^{2}**

**(ii) xy**

**Solution:**

We know that

(x + y)^{2} â€“ (x â€“ y)^{2} = 4xy

Substituting the values

6^{2}Â â€“ 4^{2} = 4xy

By further calculation

36 â€“ 16 = 4xy

20 = 4xy

4xy = 20

So we get

xy = 20/4 = 5

(i) x^{2} + y^{2} = (x + y)^{2} â€“ 2xy

Substituting the values

= 6^{2} â€“ 2 Ã— 5

By further calculation

= 36 â€“ 10

= 26

(ii) xy = 5

**10. If x â€“ 3 = 1/x, find the value of x ^{2} + 1/x^{2}.**

**Solution:**

It is given that

x â€“ 3 = 1/x

We can write it as

x â€“ 1/x = 3

Here

(x â€“ 1/x)^{2} = x^{2}Â + 1/x^{2}Â â€“ 2

So we get

x^{2}Â + 1/x^{2}Â = (x â€“ 1/x)^{2} + 2

Substituting the values

x^{2}Â + 1/x^{2} = 3^{2}Â + 2

By further calculation

= 9 + 2

= 11

**11. If x + y = 8 and xy = 3 Â¾, find the values of**

**(i) x â€“ y**

**(ii) 3 (x ^{2} + y^{2})**

**(iii) 5 (x ^{2} + y^{2}) + 4 (x â€“ y).**

**Solution:**

(i) We know that

(x â€“ y)^{2} = x^{2} + y^{2}Â â€“ 2xy

It can be written as

(x â€“ y)^{2}Â = x^{2}Â + y^{2} + 2xy â€“ 4xy

(x â€“ y)^{2}Â = (x + y)^{2}Â â€“ 4xy

It is given that

x + y = 8 and xy = 3 Â¾ = 15/4

Substituting the values

(x â€“ y)^{2}Â = 8^{2}Â â€“ 4 Ã— 15/4

So we get

(x â€“ y)^{2}Â = 65 â€“ 15 = 49

x â€“ y = Â± âˆš49 = Â± 7

(ii) We know that

(x + y)^{2} = x^{2} + y^{2}Â + 2xy

We can write it as

x^{2}Â + y^{2}Â = (x + y)^{2} – 2xy

It is given that

x + y = 8 and xy = 3 Â¾ = 15/4

Substituting the values

x^{2}Â + y^{2}Â = 8^{2}Â â€“ 2 Ã— 15/4

So we get

x^{2}Â + y^{2}Â = 64 â€“ 15/2

Taking LCM

x^{2}Â + y^{2}Â = (128 â€“ 15)/ 2 = 113/2

We get

3 (x^{2} + y^{2}) = 3 Ã— 113/2 = 339/2 = 169 Â½

(iii) We know that

5 (x^{2} + y^{2}) + 4 (x â€“ y) = 5 Ã— 113/2 + 4 Ã— Â± 7

By further calculation

= 565/2 Â± 28

We can write it as

= 565/2 + 28 or 565/2 â€“ 28

= 621/2 or 509/2

It can be written as

= 310 Â½ or 254 Â½

**12. If x ^{2} + y^{2} = 34 and xy = 10 Â½, find the value of 2 (x + y)^{2} + (x â€“ y)^{2}.**

**Solution:**

It is given that

x^{2} + y^{2} = 34 and xy = 10 Â½ = 21/2

We know that

(x + y)^{2} = x^{2} + y^{2} + 2xy

Substituting the values

(x + y)^{2} = 34 + 2 (21/2)

So we get

(x + y)^{2} = 55 â€¦.. (1)

We know that

(x â€“ y)^{2} = x^{2} + y^{2} â€“ 2xy

Substituting the values

(x â€“ y)^{2} = 34 – 2 (21/2)

So we get

(x â€“ y)^{2} = 34 â€“ 21 = 13 â€¦.. (2)

Using both the equations

2 (x + y)^{2} + (x â€“ y)^{2} = 2 Ã— 55 + 13 = 123

**13. If a â€“ b = 3 and ab = 4, find a ^{3} â€“ b^{3}.**

**Solution:**

We know that

a^{3} â€“ b^{3} = (a â€“ b)^{3} + 3ab (a + b)

Substituting the values

a^{3} â€“ b^{3} = 3^{3} + 3 Ã— 4 Ã— 3

By further calculation

a^{3} â€“ b^{3} = 27 + 36 = 63

**14. If 2a â€“ 3b = 3 and ab = 2, find the value of 8a ^{3} â€“ 27b^{3}.**

**Solution:**

We know that

8a^{3} â€“ 27b^{3} = (2a)^{3} â€“ (3b)^{3}

According to the formula

= (2a â€“ 3b)^{3} + 3 Ã— 2a Ã— 3b (2a â€“ 3b)

By further simplification

= (2a â€“ 3b)^{3} + 18ab (2a â€“ 3b)

Substituting the values

= 3^{3} + 18 Ã— 2 Ã— 3

By further calculation

= 27 + 108

= 135

**15. If x + 1/x = 4, find the values of**

**(i) x ^{2} + 1/x^{2}**

**(ii) x ^{4} + 1/x^{4}**

**(iii) x ^{3} + 1/x^{3}**

**(iv) x â€“ 1/x.**

**Solution:**

(i) We know that

(x + 1/x)^{2} = x^{2} + 1/x^{2} + 2

It can be written as

x^{2} + 1/x^{2}Â = (x + 1/x)^{2} â€“ 2

Substituting the values

= 4^{2} â€“ 2

= 16 â€“ 2

= 14

(ii) We know that

(x^{2} + 1/x^{2})^{2} = x^{4} + 1/x^{4} + 2

It can be written as

x^{4} + 1/x^{4} = (x^{2} + 1/x^{2})^{2} â€“ 2

Substituting the values

= 14^{2} â€“ 2

= 196 â€“ 2

= 194

(iii) We know that

x^{3} + 1/x^{3} = (x + 1/x)^{3} â€“ 3x (1/x) (x + 1/x)

It can be written as

(x + 1/x)^{3} â€“ 3(x + 1/x) = 4^{3} â€“ 3 Ã— 4

By further calculation

= 64 â€“ 12

= 52

(iv) We know that

(x â€“ 1/x)^{2} = x^{2} + 1/x^{2} â€“ 2

Substituting the values

= 14 â€“ 2

= 12

So we get

x â€“ 1/x = Â± 2âˆš3

**16. If x â€“ 1/x = 5, find the value of x ^{4} + 1/x^{4}.**

**Solution:**

We know that

(x â€“ 1/x)^{2} = x^{2} + 1/x^{2} â€“ 2

It can be written as

x^{2} + 1/x^{2} = (x â€“ 1/x)^{2} + 2

Substituting the values

x^{2} + 1/x^{2} = 5^{2} + 2 = 27

Here

x^{4} + 1/x^{4} = (x^{2} + 1/x^{2})^{2} â€“ 2

Substituting the values

x^{4} + 1/x^{4} = 27^{2} â€“ 2

So we get

= 729 â€“ 2

= 727

**17. If x â€“ 1/x = âˆš5, find the values of**

**(i) x ^{2} + 1/x^{2}**

**(ii) x + 1/x**

**(iii) x ^{3} + 1/x^{3}**

**Solution:**

(i) x^{2} + 1/x^{2} = (x â€“ 1/x)^{2} + 2

Substituting the values

= (âˆš5)^{2} + 2

= 5 + 2

= 7

(ii) (x + 1/x)^{2} = x^{2} + 1/x^{2} + 2

Substituting the values

= 7 + 2

= 9

Here

(x + 1/x)^{2} = 9

So we get

(x + 1/x) = Â± âˆš9 = Â± 3

(iii) x^{3} + 1/x^{3} = (x + 1/x)^{3} â€“ 3x (1/x) (x + 1/x)

Substituting the values

= (Â± 3)^{3} â€“ 3 (Â± 3)

By further calculation

= (Â± 27) â€“ (Â± 9)

= Â± 18

**18. If x + 1/x = 6, find **

**(i) x â€“ 1/x**

**(ii) x ^{2} â€“ 1/x^{2}.**

**Solution:**

(i) We know that

(x â€“ 1/x)^{2} = x^{2} + 1/x^{2} â€“ 2

It can be written as

(x â€“ 1/x)^{2} = x^{2} + 1/x^{2} + 2 â€“ 4

(x â€“ 1/x)^{2} = (x + 1/x)^{2} â€“ 4

Substituting the values

(x â€“ 1/x)^{2} = 6^{2} â€“ 4 = 32

So we get

x â€“ 1/x = Â± âˆš32 = Â± 4âˆš2

(ii) We know that

x^{2}Â â€“ 1/x^{2} = (x â€“ 1/x) (x + 1/x)

Substituting the values

x^{2}Â â€“ 1/x^{2} = (Â± 4âˆš2) (6) = Â± 24 âˆš2

**19. If x + 1/x = 2, prove that x ^{2} + 1/x^{2} = x^{3}Â + 1/x^{3} = x^{4} + 1/x^{4}.**

**Solution:**

We know that

x^{2} + 1/x^{2} = (x + 1/x) â€“ 2

Substituting the values

x^{2} + 1/x^{2} = 2^{2} â€“ 2

So we get

x^{2} + 1/x^{2} = 4 â€“ 2 = 2 â€¦. (1)

x^{3}Â + 1/x^{3} = (x + 1/x)^{3} â€“ 3 (x + 1/x)

Substituting the values

x^{3}Â + 1/x^{3} = 2^{3} â€“ 3 Ã— 2

So we get

x^{3}Â + 1/x^{3} = 8 â€“ 6 = 2 â€¦â€¦ (2)

x^{4} + 1/x^{4} = (x^{2} + 1/x^{2})^{2}Â â€“ 2

Substituting the values

x^{4} + 1/x^{4}Â = 2^{2}Â â€“ 2

So we get

x^{4} + 1/x^{4} = 4 â€“ 2 = 2 â€¦. (3)

From equation (1), (2) and (3)

x^{2} + 1/x^{2} = x^{3}Â + 1/x^{3} = x^{4} + 1/x^{4}

Hence, it is proved.

**20. If x â€“ 2/x = 3, find the value of x ^{3} â€“ 8/x^{3}.**

**Solution:**

We know that

(x â€“ 2/x)^{3} = x^{3} â€“ 8/x^{3} â€“ 3 (x) (2/x) (x â€“ 2/x)

By further simplification

(x â€“ 2/x)^{3} = x^{3} â€“ 8/x^{3} â€“ 6 (x â€“ 2/x)

It can be written as

x^{3} â€“ 8/x^{3} = (x â€“ 2/x)^{3} + 6 (x â€“ 2/x)

Substituting the values

x^{3} â€“ 8/x^{3} = 3^{3} + 6 Ã— 3

By further calculation

x^{3} â€“ 8/x^{3} = 27 + 18 = 45

**21. If a + 2b = 5, prove that a ^{3} + 8b^{3} + 30ab = 125.**

**Solution:**

We know that

(a + 2b)^{3}Â = a^{3} + 8b^{3} + 3 (a) (2b) (a + 2b)

Substituting the values

5^{3} = a^{3} + 8b^{3} + 6ab (5)

By further calculation

125 = a^{3} + 8b^{3} + 30ab

Therefore, a^{3} + 8b^{3} + 30ab = 125.

**22. If a + 1/a = p, prove that a ^{3} + 1/a^{3} = p (p^{2} â€“ 3).**

**Solution:**

We know that

a^{3} + 1/a^{3} = (a + 1/a)^{3} â€“ 3a (1/a) (a + 1/a)

Substituting the values

a^{3} + 1/a^{3} = p^{3} â€“ 3p

Taking p as common

a^{3} + 1/a^{3} = p (p^{2}Â â€“ 3)

Therefore, it is proved.

**23. If x ^{2} + 1/x^{2} = 27, find the value of x â€“ 1/x.**

**Solution:**

We know that

(x â€“ 1/x)^{2} = x^{2} + 1/x^{2} â€“ 2

Substituting the values

(x â€“ 1/x)^{2} = 27 â€“ 2 = 25

So we get

x â€“ 1/x = Â± âˆš25 = Â± 5

**24. If x ^{2} + 1/x^{2} = 27, find the value of 3x^{3} + 5x â€“ 3/x^{3} â€“ 5/x.**

**Solution:**

We know that

(x â€“ 1/x)^{2} = x^{2} + 1/x^{2} â€“ 2

Substituting the values

(x â€“ 1/x)^{2} = 27 â€“ 2 = 25

So we get

x â€“ 1/x = Â± âˆš25 = Â± 5

Here

3x^{3} + 5x â€“ 3/x^{3} â€“ 5/x = 3 (x^{3} â€“ 1/x^{3}) + 5 (x â€“ 1/x)

It can be written as

= 3 [(x â€“ 1/x)^{3} + 3 (x â€“ 1/x)] + 5 (x â€“ 1/x)

Substituting the values

= 3 [(Â± 5)^{3} + 3 (Â± 5)] + 5 (Â± 5)

By further calculation

= 3 [(Â± 125) + (Â± 15)] + (Â± 25)

So we get

= (Â± 375) + (Â± 45) + (Â± 25)

= Â± 445

**25. If x ^{2} + 1/25x^{2} = 8 3/5, find x + 1/5x.**

**Solution:**

We know that

(x + 1/5x)^{2} = x^{2} + 1/25x^{2}Â + 2x (1/5x)

It can be written as

(x + 1/5x)^{2} = x^{2} + 1/25x^{2}Â + 2/5

Substituting the values

(x + 1/5x)^{2} = 8 3/5 + 2/5

(x + 1/5x)^{2} = 43/5 + 2/5

So we get

(x + 1/5x)^{2} = 45/5 = 9

Here

x + 1/5x = Â± âˆš9 = Â± 3

**26. If x ^{2}Â + 1/4x^{2} = 8, find x^{3} + 1/8x^{3}.**

**Solution:**

We know that

(x + 1/2x)^{2}Â = x^{2} + (1/2x)^{2} + 2x (1/2x)

It can be written as

(x + 1/2x)^{2}Â = x^{2} + 1/4x^{2} + 1

Substituting the values

(x + 1/2x)^{2}Â = 8 + 1 = 9

So we get

x + 1/2x = Â± âˆš9 = Â± 3

Here

x^{3} + 1/8x^{3} = x^{3} + (1/2x)^{3}

We know that

x^{3} + 1/8x^{3} = (x + 1/2x)^{3} â€“ 3x (1/2x) (x + 1/2x)

Substituting the values

x^{3} + 1/8x^{3} = (Â± 3)^{3} â€“ 3/2 (Â± 3)

By further calculation

x^{3} + 1/8x^{3} = Â± (27 â€“ 9/2)

Taking LCM

x^{3} + 1/8x^{3} = Â± (54 â€“ 9)/ 2

x^{3} + 1/8x^{3} = Â± 45/2 = Â± 22 Â½

Therefore, x^{3} + 1/8x^{3} = Â± 22 Â½.

**27. If a ^{2} â€“ 3a + 1 = 0, find**

**(i) a ^{2} + 1/a^{2}**

**(ii) a ^{3} + 1/a^{3}.**

**Solution:**

It is given that

a^{2} â€“ 3a + 1 = 0

By dividing each term by a

a + 1/a = 3

(i) We know that

(a + 1/a)^{2} = a^{2} + 1/a^{2} + 2

It can be written as

a^{2} + 1/a^{2} = (a + 1/a)^{2} â€“ 2

Substituting the values

= 3^{2} â€“ 2

= 9 â€“ 2

= 7

(ii) We know that

(a + 1/a)^{3} = a^{3} + 1/a^{3} + 3 (a + 1/a)

It can be written as

a^{3} + 1/a^{3} = (a + 1/a)^{3} â€“ 3 (a + 1/a)

Substituting the values

= 3^{3}Â â€“ 3 (3)

= 27 â€“ 9

= 18

**28. If a = 1/ (a â€“ 5), find **

**(i) a â€“ 1/a**

**(ii) a + 1/a**

**(iii) a ^{2} â€“ 1/a^{2}.**

**Solution:**

It is given that

a = 1/ (a â€“ 5)

We can write it as

a^{2} â€“ 5a â€“ 1 = 0

Now divide each term by a

a â€“ 5 â€“ 1/a = 0

So we get

a â€“ 1/a = 5

(i) a â€“ 1/a = 5

(ii) We know that

(a + 1/a)^{2} = (a â€“ 1/a)^{2} + 4

Substituting the values

(a + 1/a)^{2} = 5^{2} + 4

So we get

(a + 1/a)^{2} = 25 + 4 = 29

a + 1/a = Â± âˆš29

(ii) We know that

a^{2} â€“ 1/a^{2} = (a + 1/a) (a â€“ 1/a)

Substituting the values

a^{2} â€“ 1/a^{2} = Â± âˆš29 Ã— 5

a^{2} â€“ 1/a^{2} = Â± 5âˆš29

**29. If (x + 1/x) ^{2} = 3, find x^{3} + 1/x^{3}.**

**Solution:**

It is given that

(x + 1/x)^{2} = 3

(x + 1/x) = Â± âˆš3

We know that

x^{3} + 1/x^{3} = (x + 1/x)^{3} â€“ 3 (x + 1/x)

Substituting the values

x^{3} + 1/x^{3} = (Â± âˆš3)^{3} â€“ 3 (Â± âˆš3)

By further calculation

x^{3} + 1/x^{3} = (Â± 3âˆš3) â€“ (Â± 3âˆš3) = 0

**30. If x = 5 – 2âˆš6, find the value of âˆšx + 1/âˆšx.**

**Solution:**

It is given that

x = 5 – 2âˆš6

We can write it as

= 5 + 2âˆš6

Here

x + 1/x = 5 – 2âˆš6 + 5 + 2âˆš6 = 10

So we get

(âˆšx + 1/âˆšx)^{2} = x + 1/x + 2

Substituting the values

= 10 + 2

= 12

**31. If a + b + c = 12 and ab + bc + ca = 22, find a ^{2} + b^{2} + c^{2}.**

**Solution:**

We know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

We can write it as

a^{2} + b^{2} + c^{2} = (a + b + c)^{2} â€“ 2 (ab + bc + ca)

Substituting the values

a^{2} + b^{2} + c^{2} = 12^{2} â€“ 2 (22)

By further calculation

a^{2} + b^{2} + c^{2} = 144 â€“ 44 = 100

**32. If a + b + c = 12 and a ^{2} + b^{2} + c^{2} = 100, find ab + bc + ca.**

**Solution:**

We know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

It can be written as

2ab + 2bc + 2ca = (a + b + c)^{2} â€“ (a^{2} + b^{2} + c^{2})

Taking out 2 as common

2 (ab + bc + ca) = 12^{2} â€“ 100 = 144 â€“ 100 = 44

By further calculation

ab + bc + ca = 44/2 = 22

**33. If a ^{2} + b^{2} + c^{2} = 125 and ab + bc + ca = 50, find a + b + c.**

**Solution:**

We know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

Substituting the values

(a + b + c)^{2} = 125 + 2 (50)

By further calculation

(a + b + c)^{2} = 125 + 100 = 225

So we get

a + b + c = Â± âˆš225 = Â± 15

**34. If a + b â€“ c = 5 and a ^{2} + b^{2} + c^{2} = 29, find the value of ab â€“ bc â€“ ca.**

**Solution:**

It is given that

a + b â€“ c = 5

By squaring on both sides

(a + b â€“ c)^{2}Â = 5^{2}

Expanding using formula

a^{2} + b^{2} + c^{2}Â + 2ab â€“ 2bc â€“ 2ca = 25

Substituting the values and taking 2 as common

29 + 2 (ab â€“ bc â€“ ca) = 25

By further calculation

2 (ab â€“ bc â€“ ca) = 25 â€“ 29 = – 4

So we get

ab â€“ bc â€“ ca = – 4/2 = – 2

Therefore, ab â€“ bc â€“ ca = – 2.

**35. If a â€“ b = 7 and a ^{2} + b^{2} = 85, then find the value of a^{3} â€“ b^{3}.**

**Solution:**

We know that

(a â€“ b)^{2} = a^{2} + b^{2} â€“ 2ab

Substituting the values

7^{2} = 85 â€“ 2ab

By further calculation

49 = 85 â€“ 2ab

So we get

2ab = 85 â€“ 49 = 36

Dividing by 2

ab = 36/2 = 18

Here

a^{3}Â â€“ b^{3} = (a â€“ b) (a^{2} + b^{2}Â + ab)

Substituting the values

a^{3}Â â€“ b^{3} = 7 (85 + 18)

By further calculation

a^{3}Â â€“ b^{3} = 7 Ã— 103

So we get

a^{3}Â â€“ b^{3} = 721

**36. If the number x is 3 less than the number y and the sum of the squares of x and y is 29, find the product of x and y.**

**Solution:**

It is given that

x = y – 3 and x^{2}Â + y^{2} = 29

It can be written as

x â€“ y = – 3

By squaring on both sides

(x â€“ y)^{2} = (-3)^{2}

Expanding using formula

x^{2} + y^{2} â€“ 2xy = 9

Substituting the values

29 â€“ 2xy = 9

By further calculation

-2xy = 9 â€“ 29 = – 20

Dividing by 2

xy = – 20/-2 = 10

So we get

xy = 10

**37. If the sum and the product of two numbers are 8 and 15 respectively, find the sum of their cubes.**

**Solution:**

Consider x and y as the two numbers

x + y = 8 and xy = 15

By cubing on both sides

(x + y)^{3} = 8^{3}

Expanding using formula

x^{3} + y^{3} + 3xy (x + y) = 512

Substituting the values

x^{3} + y^{3} + 3 Ã— 15 Ã— 8 = 512

By further calculation

x^{3} + y^{3} + 360 = 512

So we get

x^{3} + y^{3} = 512 â€“ 360 = 152

Chapter Test

**1. Find the expansions of the following :(i) (2x + 3y + 5) (2x + 3y â€“ 5)(ii) (6 â€“ 4a -7b)**

^{2}(iii) (7 â€“ 3xy)

^{3}(iv) (x + y + 2)

^{3}

**Solution:**

**(i)** (2x + 3y + 5) (2x + 3y â€“ 5)

Let us simplify the expression, we get

(2x + 3y + 5) (2x + 3y â€“ 5) = [(2x + 3y) + 5] [(2x – 3y) â€“ 5]

By using the formula, (a)^{2} â€“ (b)^{2} = [(a + b) (a – b)]

= (2x + 3y)^{2}Â â€“ (5)^{2}

= (2x)^{2}Â + (3y)Â ^{2}Â + 2 Ã— 2x Ã— 3y â€“ 5 Ã— 5

= 4x^{2}Â + 9y^{2}Â + 12xy – 25

Â

**(ii)** (6 â€“ 4a â€“ 7 b)^{2}

Let us simplify the expression, we get

(6 â€“ 4a â€“ 7 b)^{2}Â = [ 6 + (- 4a) + (-7b)]^{2}

= (6)^{2}Â + (- 4a)^{2}Â + (- 7b)^{2}Â + 2 (6) (- 4a) + 2 (- 4a) (-7b) + 2 (-7b) (6)

= 36 + 16a^{2}Â + 49b^{2}Â â€“ 48a + 56ab â€“ 84b

**(iii)** (7 â€“ 3xy)^{3}

Let us simplify the expression

By using the formula, we get

(7 â€“ 3xy)^{3 }= (7)^{3}Â â€“ (3xy)^{3}Â â€“ 3 (7) (3xy) (7 â€“ 3xy)

Â = 343 â€“ 27x^{3}y^{3}Â â€“ 63xy (7 â€“ 3xy)

= 343 â€“ 27x^{3}y^{3}Â â€“ 441xy + 189x^{2}y^{2}

**(iv)** (x + y + 2)^{3}Â

Let us simplify the expression

By using the formula, we get

(x + y + 2 )^{3}Â = [(x + y) + 2]^{3}

= (x + y)^{3}Â + (2)^{3}Â + 3 (x + y) (2) (x + y + 2)

= x^{3}Â + y^{3}Â + 3x^{2}y + 3xy^{2}Â + 8 + 6 (x + y) [(x + y) + 2]

= x^{3}Â + y^{3}Â + 3x^{2}y + 3xy^{2}Â + 8 + 6 (x + y)^{2}Â +12(x + y)

= x^{3}Â + y^{3}Â + 3x^{2}y + 3xy^{2}Â + 8 + 6 (x^{2}Â + y^{2}Â + 2xy) + 12x + 12y = x^{3}Â + y^{3}Â + 3x^{2}y + 3xy^{2}Â + 8 + 6x^{2}Â + 6y^{2}Â + 12xy + 12x + 12y

= x^{3}Â + y^{3}Â + 3x^{2}y + 3xy^{2}Â + 8 + 6x^{2}Â + 6y^{2}Â + 12x + 12y + 12xy

**2. Simplify: (x â€“ 2) (x + 2) (x ^{2}Â + 4) (x^{4}Â + 16)**

Solution:

Let us simplify the expression, we get

(x â€“ 2) (x + 2) (x^{4}Â + 4) (x^{4}Â +16) = (x^{2}Â â€“ 4) (x^{4}Â + 4) (x^{4}Â + 16)

= [(x^{2})^{2}Â â€“ (4)^{2}] (x^{4}Â + 16)

= (x^{4}Â â€“ 16) (x^{4}Â + 16)

= (x^{4})^{2}Â â€“ (16)^{2}Â

= x^{8}Â â€“ 256

**3. Evaluate 1002 Ã— 998 by using a special product.Solution: **

Let us simplify the expression, we get

1002 Ã— 998 = (1000 + 2) (1000 â€“ 2)

= (1000)^{2}Â â€“ (2)^{2}Â

= 1000000 â€“ 4

= 999996

**4. If a + 2b + 3c = 0, Prove that a ^{3}Â + 8b^{3}Â + 27c^{3}Â = 18 abc**

Solution:

Given:

a + 2b + 3c = 0, a + 2b = â€“ 3c

Let us cube on both the sides, we get

(a + 2b)^{3} = (-3c)^{3}

a^{3}Â + (2b)^{3}Â + 3(a) (2b) (a + 2b) = -27c^{3}

a^{3}Â + 8b^{3}Â + 6ab (â€“ 3c) = â€“ 27c^{3}

a^{3}Â + 8b^{3}Â â€“ 18abc = -27c^{3}

a^{3}Â + 8b^{3}Â + 27c^{3}Â = 18abc

Hence proved.

**5. If 2x = 3y â€“ 5, then find the value of 8x ^{3}Â â€“ 27y^{3}Â + 90xy + 125.**

Solution:

Given:

2x = 3y â€“ 5

2x â€“ 3y = -5

Now, let us cube on both sides, we get

(2x â€“ 3y)^{3}Â = (-5)^{3}

(2x)^{3}Â â€“ (3y)^{3}Â â€“ 3 Ã— 2x Ã— 3y (2x â€“ 3y) = -125

8x^{3}Â â€“ 27y^{3}Â â€“ 18xy (2x â€“ 3y) = -125

Now, substitute the value of 2x – 3y = -5

8x^{3}Â â€“ 27y^{3}Â â€“ 18xy (-5) = -125

8x^{3}Â â€“ 27y^{3}Â + 90xy = -125

8x^{3}Â â€“ 27y^{3}Â + 90xy + 125 = 0

**6. If a ^{2} â€“ 1/a^{2} = 5, evaluate a^{4} + 1/a^{4}**

**Solution:**

It is given that,

a^{2} â€“ 1/a^{2} = 5

So,

By using the formula, (a + b)^{2}

^{2}â€“ 1/a

^{2}]

^{2}= a

^{4}+ 1/a

^{4}â€“ 2 [a

^{2}â€“ 1/a

^{2}]

^{2}+ 2 = a

^{4}+ 1/a

^{4}

Substitute the value of a^{2} â€“ 1/a^{2} = 5, we get

5^{2} + 2 = a^{4} + 1/a^{4}

a^{4} + 1/a^{4} = 25 + 2

= 27

**7. If a + 1/a = p and a â€“ 1/a = q, Find the relation between p and q.**

**Solution:**

It is given that,

a + 1/a = p and a â€“ 1/a = q

so,

(a + 1/a)^{2} â€“ (a â€“ 1/a)^{2} = 4(a) (1/a)

= 4

By substituting the values, we get

p^{2} â€“ q^{2} = 4

Hence the relation between p and q is that p^{2} â€“ q^{2} = 4.

**8. If (a ^{2} + 1)/a = 4, find the value of 2a^{3} + 2/a^{3}**

**Solution:**

It is given that,

(a^{2} + 1)/a = 4

a^{2}/a + 1/a = 4

a + 1/a = 4

So by multiplying the expression by 2a, we get

2a^{3} + 2/a^{3} = 2[a^{3} + 1/a^{3}]

= 2 [(a + 1/a)^{3} â€“ 3 (a) (1/a) (a + 1/a)]

= 2 [(4)^{3} â€“ 3(4)]

= 2 [64 – 12]

= 2 (52)

= 104

**9. If x = 1/(4 – x), find the value of**

**(i) x + 1/x**

**(ii) x ^{3} + 1/x^{3} **

**(iii) x ^{6} + 1/x^{6} **

**Solution:**

It is given that,

x = 1/(4 – x)

So,

**(i)** x(4 – x) = 1

4x â€“ x^{2} = 1

Now let us divide both sides by x, we get

4 â€“ x = 1/x

4 = 1/x + x

1/x + x = 4

1/x + x = 4

**(ii)** x^{3} + 1/x^{3} = (x + 1/x)^{2} â€“ 3(x + 1/x)

By substituting the values, we get

= (4)^{3} â€“ 3(4)

= 64 â€“ 12

= 52

**(iii)** x^{6} + 1/x^{6} = (x^{3} + 1/x^{3})^{2} â€“ 2

= (52)^{2} â€“ 2

= 2704 â€“ 2

= 2702

**10. If x â€“ 1/x = 3 + 2âˆš2, find the value of Â¼ (x ^{3} â€“ 1/x^{3})**

**Solution:**

It is given that,

x â€“ 1/x = 3 + 2âˆš2

So,

x^{3} â€“ 1/x^{3} = (x â€“ 1/x)^{3} + 3(x â€“ 1/x)

= (3 + 2âˆš2)^{3} + 3(3 + 2âˆš2)

By using the formula, (a+b)^{3} = a^{3} + b^{3} + 3ab (a + b)

= (3)^{3} + (2âˆš2)^{3} + 3 (3) (2âˆš2) (3 + 2âˆš2) + 3(3 + 2âˆš2)

= 27 + 16âˆš2 + 54âˆš2 + 72 + 9 + 6âˆš2

= 108 + 76âˆš2

Hence,

Â¼ (x^{3} â€“ 1/x^{3}) = Â¼ (108 + 76âˆš2)

= 27 + 19âˆš2

**11. If x + 1/x = 3 1/3, find the value of x ^{3} â€“ 1/x^{3}**

**Solution:**

It is given that,

x + 1/x = 3 1/3

we know that,

(x â€“ 1/x)^{2} = x^{2} + 1/x^{2} â€“ 2

= x^{2} + 1/x^{2} + 2 â€“ 4

= (x + 1/x)^{2} â€“ 4

But x + 1/x = 3 1/3 = 10/3

So,

(x â€“ 1/x)^{2} = (10/3)^{2} â€“ 4

= 100/9 â€“ 4

= (100 – 36)/9

= 64/9

x â€“ 1/x = âˆš(64/9)

= 8/3

Now,

x^{3} â€“ 1/x^{3} = (x â€“ 1/x)^{3} + 3 (x) (1/x) (x â€“ 1/x)

= (8/3)^{3} + 3 (8/3)

= ((512/27) + 8)

= 728/27

= 26 26/27

**12. If x = 2 – âˆš3, then find the value of x ^{3} â€“ 1/x^{3}**

**Solution:**

It is given that,

x = 2 – âˆš3

so,

1/x = 1/(2 – âˆš3)

By rationalizing the denominator, we get

= [1(2 + âˆš3)] / [(2 – âˆš3) (2 + âˆš3)]

= [(2 + âˆš3)] / [(2^{2}) â€“ (âˆš3)^{2}]

= [(2 + âˆš3)] / [4 – 3]

= 2 + âˆš3

Now,

x â€“ 1/x = 2 – âˆš3 â€“ 2 – âˆš3

= – 2âˆš3

Let us cube on both sides, we get

(x â€“ 1/x)^{3} = (-2âˆš3)^{3}

x^{3} â€“ 1/x^{3} â€“ 3 (x) (1/x) (x â€“ 1/x) = 24âˆš3

x^{3} â€“ 1/x^{3} â€“ 3(-2âˆš3) = -24âˆš3

x^{3} â€“ 1/x^{3} + 6âˆš3 = -24âˆš3

x^{3} â€“ 1/x^{3} = -24âˆš3 – 6âˆš3

= -30âˆš3

Hence,

x^{3} â€“ 1/x^{3} = -30âˆš3

**13. If the sum of two numbers is 11 and sum of their cubes is 737, find the sum of their squares.Solution: **

Let us consider x and y be two numbers

Then,

x + y = 11

x^{3}Â + y^{3}Â = 735 and x^{2}Â + y^{2}Â =?

Now,

x + y = 11

Let us cube on both the sides,

(x + y)^{3}Â = (11)^{3}

x^{3}Â + y^{3}Â + 3xy (x + y) = 1331

737 + 3x Ã— 11 = 1331

33xy = 1331 â€“ 737

= 594

xy = 594/33

xy =Â 8

We know that, x + y = 11

By squaring on both sides, we get

(x + y)^{2}Â = (11)^{2}

x^{2}Â + y^{2}Â + 2xy = 121Â ^{2}Â x^{2}Â + y^{2}Â + 2 Ã— 18 = 121

x^{2}Â + y^{2}Â + 36 = 121

x^{2}Â + y^{2}Â = 121 â€“ 36

= 85

Hence sum of the squares = 85

**14. If a â€“ b = 7 and a ^{3}Â â€“ b^{3}Â = 133, find:**

(i) ab

(ii) a^{2}Â + b^{2}

Solution:

It is given that,

a â€“ b = 7

let us cube on both sides, we get

**(i)** (a â€“ b)^{3}Â = (7)^{3}

a^{3}Â + b^{3}Â â€“ 3ab (a â€“ b) = 343

133 â€“ 3ab Ã— 7 = 343

133 â€“ 21ab = 343

â€“ 21ab = 343 â€“ 133 21ab

= 210

ab = -210/21

ab = -10

**(ii)** a^{2}Â + b^{2}

Again a â€“ b = 7

Let us square on both sides, we get

(a â€“ b)^{2}Â = (7)^{2}

a^{2}Â + b^{2}Â â€“ 2ab = 49

a^{2}Â + b^{2}Â â€“ 2 Ã— (- 10) = 49

a^{2}Â + b^{2}Â + 20 = 49

a^{2}Â + b^{2}Â = 49 â€“ 20

= 29

Hence, a^{2}Â + b^{2}Â = 29

**15. Find the coefficient of x ^{2}Â expansion of (x^{2}Â + x + 1)^{2}Â + (x^{2}Â â€“ x + 1)^{2}**

Solution:

Given:

The expression, (x^{2}Â + x + 1)^{2}Â + (x^{2}Â â€“ x + 1)^{2}

(x^{2}Â + x + 1)^{2}Â + (x^{2}Â â€“ x + 1)^{2} = [((x^{2}Â + 1) + x)^{2}Â + [(x^{2}Â + 1) â€“ x)^{2}]

= (x^{2}Â + 1)^{2}Â + x^{2}Â + 2 (x^{2}Â + 1) (x) + (x^{2}Â + 1)^{2}Â + x^{2}Â â€“ 2 (x^{2}Â + 1) (x)

= (x^{2})^{2}Â + (1)^{2}Â + 2 Ã— x^{2}Â Ã— 1 + x^{2}Â + (x^{2})^{2}Â + 1 + 2 Ã— x^{2}Â + 1 + x^{2}

= x^{4}Â + 1 + 2x^{2} + x^{2}Â + x^{4}Â + 1 + 2x^{2}Â + x^{2}

= 2x^{4}Â + 6x^{2}Â + 2

âˆ´ Co-efficient of x^{2}Â is 6.