RS Aggarwal Solutions For Class 8 Maths Chapter 6 Operations on Algebraic Expressions are provided here. You can download the pdf of RS Aggarwal Solutions for Class 8 Maths Chapter 6 Operations on Algebraic Expression from the given links. Class 8 is an important phase of a studentâ€™s life. It is crucial concepts that are taught in Class 8 are understood thoroughly as these concepts are continued in Class 9 and 10 and are also the foundation for higher studies. Here we will learn about the basic operations that are addition, subtraction, multiplication and division on algebraic expressions.

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## Exercise 6A

**Add:**

**1. 8ab, -5ab, 3ab, -ab**

**2. 7x, -3x, 5x, -x, -3x**

**3. 3a-3b+4c, 2a+3b-8c, a-6b+c**

**4. 5x-8y+2z, 3z-4y-2x, 6y-z-x and 3x-2z-3y**

**5. 6ax-2by+3cz, 6by-11ax-cz and 10cz-2ax-3by**

**6. 2x ^{3}-9x^{2}+8, 3x^{2}-6x-5, 7x^{3}-10x+1 and 3+2x-5x^{2}-4x^{3}**

**7. 6p+4q-r+3, 2r-5p-6, 11q-7p+2r-1 and 2q-3r+4**

**8. 4x ^{2}-7xy+4y2-3, 5+6y2-8xy+x^{2} and 6-2xy+2x^{2}-5y2**

**Solution:**

1. Given 8ab, -5ab, 3ab, -ab

To add the given expression we have arrange them column wise is given below:

8 ab

-5 ab

3 ab

-ab

5ab |

2. Given 7x,-3x, 5x, -x, -2x

To add the given expression we have arrange them column wise is given below:

7x

-3x

5x

-x

-2x

6x |

3. Given 3a-3b+4c, 2a+3b-8c, a-6b+c

To add the given expression we have arrange them column wise is given below:

3a-3b+4c

2a+3b-8c

a-6b+c

6a-6b-3c |

4. Given 5x-8y+2z, 3z-4y-2x, 6y-z-x and 3x-2z-3y

To add the given expression we have arrange them column wise is given below:

5x-8y+2z

-2x-4y+3z

-x+6y-z

3x – 2z – 3y

5x-9y+2z |

5. Given 6ax-2by+3cz, 6by-11ax-cz and 10cz-2ax-3by

To add the given expression we have arrange them column wise is given below:

6ax-2by+3cz

-11ax+6by-cz

-2ax-3by+10cz

-7ax+by+12cz |

6. Given 2x^{3}-9x^{2}+8, 3x^{2}-6x-5, 7x^{3}-10x+1 and 3+2x-5x^{2}-4x^{3}

To add the given expression we have arrange them column wise is given below:

2x^{3}-9x^{2}+8

7x^{3}-10x+1

3x^{2}– 6x- 5

-4x^{3}-5x^{2}+2x+3

5x^{3}-11x^{2}-14x+7 |

7. Given 6p+4q-r+3, 2r-5p-6, 11q-7p+2r-1 and 2q-3r+4

To add the given expression we have arrange them column wise is given below:

6p+4q-r+3

-7p+11q+2r-1

-5p+2r-6

2q-3r+4

-6p+17q |

8. Given 4x^{2}-7xy+4y2-3, 5+6y2-8xy+x^{2} and 6-2xy+2x^{2}-5y2

To add the given expression we have arrange them column wise is given below:

4x^{2}-7xy+4y2-3

x^{2}-8xy+6y2+5

2x^{2}-2xy-5y2+6

7x^{2}+5y2-17xy+8 |

**Subtract: **

**9. 3ax ^{2}b from -5ax^{2}b**

**10. -8pq from 6pq**

**11.-2abc from -8abc**

**12.-16p from -11p**

**13. 2a-5b+2c-9 from 3a-4b-c+6**

Solution:

9. Given 3ax^{2}b from -5ax^{2}b

According to the rules of subtraction of algebraic equations, we have both expressions with negative sign so we have to add the expressions.

Now arrange the variables in rows and columns we get

-5ax^{2}b

3ax^{2}b

–

-8 ax^{2}b |

And we have to keep big numerical sign

10. Given -8pq from 6pq

According to the rules of subtraction of algebraic equations, we have negative sign will becomes positive and so we have to keep the big numerical sign.

Now arrange the variables in rows and columns we get

6pq

– 8pq

+

+14 pq |

11. Given -2abc from -8abc

According to the rules of subtraction of algebraic equations, we have negative sign will becomes positive and so we have to keep the big numerical sign.

Now arrange the variables in rows and columns we get

– 8abc – (-2abc) = – 8abc + 2abc = – 6abc

-8abc

-2abc

+

– 6 abc |

12. Given -16p from -11p

According to the rules of subtraction of algebraic equations, we have negative sign will becomes positive and so we have to keep the big numerical sign.

Now arrange the variables in rows and columns we get

– 11p – (-16p) = – 11p + 16p = 5p

-16p

-11p

+

+ 5p |

13. Given 2a-5b+2c-9 from 3a-4b-c+6

Now arrange the variables in rows and columns we get

3a-4b-c+6

2a-5b+2c-9

– +- +

+ a+b-3c+15 |

## Exercise 6B

**Find each of the following products:**

**1. (5x + 7) Ã— (3x + 4)**

**2. (4x + 9) Ã— (x â€“ 6)**

**3. (2x +5 ) Ã— (4x â€“ 3)**

**4. (3y â€“ 8) Ã— (5y â€“ 1)**

**5. (7x + 2y) Ã— (x + 4y)**

**6. (9x + 5y) Ã— (4x + 3y)**

**7. (3m â€“ 4n) Ã— (2m â€“ 3n)**

**8. (x ^{2}– a^{2}) Ã— (x-a)**

**9. (x ^{2}-y^{2}) Ã— (x + 2y)**

**10. (3p ^{2}+q^{2}) Ã— (x^{2}-y^{2})**

**11. (2x ^{2}-5y^{2}) Ã— (x^{2}+3y^{2})**

**12. (x ^{3}-y^{3}) Ã— (x^{2} + y^{2})**

**13. (x ^{4}+y^{4}) Ã— (x^{2} â€“y^{2})**

**14. (x ^{4}+(1/x^{4}) Ã— ( x + (1/x))**

**Solution:**

1. Given (5x + 7) Ã— (3x + 4)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(5x + 7) Ã— (3x + 4)

â‡’5x (3x + 4) + 7 (3x + 4)

â‡’15x^{2} + 20x + 21x + 28

â‡’ 15x^{2} + 41x + 28

2. Given (4x + 9) Ã— (x â€“ 6)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(4x + 9) Ã— (x â€“ 6)

â‡’4x (x – 6) + 9 (x – 6)

â‡’4x^{2} – 24x + 9x – 54

â‡’ 4x^{2} – 15x â€“ 54

3. Given (2x + 5) Ã— (4x â€“ 3)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(2x + 5) Ã— (4x â€“ 3)

â‡’2x (4x – 3) + 5 (4x – 3)

â‡’8x^{2} – 6x + 20x – 15

â‡’ 8x^{2} + 14x – 15

4. Given (3y â€“ 8) Ã— (5y â€“ 1)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(3y â€“ 8) Ã— (5y â€“ 1)

â‡’3y (5y – 1) – 8 (5y – 1)

â‡’15y^{2} â€“ 3y â€“ 40y + 8

â‡’15y^{2} â€“ 43y + 8

5. Given (7x + 2y) Ã— (x + 4y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(7x + 2y) Ã— (x + 4y)

â‡’7x (x + 4y) + 2y (x + 4y)

â‡’7x^{2} + 28xy + 2yx + 8y^{2}

â‡’ 7x^{2} + 30xy + 8y^{2}

6. Given (9x + 5y) Ã— (4x + 3y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(9x + 5y) Ã— (4x + 3y)

â‡’9x (4x + 3y) + 5y (4x + 3y)

â‡’36x^{2} + 27xy + 20yx + 15y^{2}

â‡’ 36x^{2} + 47x + 15 y^{2}

7. Given (3m â€“ 4n) Ã— (2m â€“ 3n)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(3m â€“ 4n) Ã— (2m â€“ 3n)

â‡’3m (2m â€“ 3n) â€“ 4n (2m â€“ 3n)

â‡’6m^{2} – 9mn â€“ 8mn + 12n^{2}

â‡’ 6m^{2} â€“ 17mn + 12 n^{2}

8. Given (x^{2}– a2) Ã— (x-a)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x^{2}– a^{2}) Ã— (x-a)

â‡’ x^{2}(x-a) – a^{2}(x-a)

â‡’ x^{3}-ax^{2}-a^{2}x+a^{3}

9. Given (x^{2}-y^{2}) Ã— (x + 2y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x^{2}-y^{2}) Ã— (x + 2y)

â‡’ x^{2}(x + 2y) â€“ y^{2}(x + 2y)

â‡’ x^{3} + 2x^{2}y â€“ xy^{2} â€“ 2y^{3}

10. Given (3p^{2}+q^{2}) Ã— (2p^{2}-3q^{2})

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(3p^{2}+q^{2}) Ã— (2p^{2}-3q^{2})

â‡’ 3p^{2} (2p^{2}-3q^{2}) + q^{2} (2p^{2}-3q^{2})

â‡’ 6p^{4} â€“ 7p^{2}q^{2} â€“ 3q^{4}

11. Given (2x^{2}-5y^{2}) Ã— (x^{2}+3y^{2})

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(2x^{2}-5y^{2}) Ã— (x^{2}+3y^{2})

â‡’ 2x^{2} (x^{2}+3y^{2}) – 5y^{2}(x^{2}+3y^{2})

â‡’ 2x^{4} + x^{2}y^{2} â€“ 15y^{4}

12. Given (x^{3}-y^{3}) Ã— (x^{2} + y^{2})

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x^{3}-y^{3}) Ã— (x^{2} + y^{2})

â‡’ x^{3} (x^{2} + y^{2}) -y^{3}(x^{2} + y^{2})

â‡’ x^{5} + x^{3}y^{2} â€“ x^{2}y^{3} – y^{5}

13. Given (x^{4}+y^{4}) Ã— (x^{2} â€“y^{2})

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x^{4}+y^{4}) Ã— (x^{2} â€“y^{2})

â‡’ x^{4} (x^{2} â€“y^{2}) + y^{4}(x^{2} â€“y^{2})

â‡’ x^{6} â€“ x^{4}y^{2} + x^{2}y^{4} â€“ y^{6}

14. Give. (x^{4}+(1/x^{4}) Ã— ( x + (1/x))

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x^{4}+(1/x^{4}) Ã— ( x + (1/x))

â‡’ x^{4}( x + (1/x)) + (1/x^{4}) ( x + (1/x))

â‡’ x^{5}+x^{3}+(1/x^{3})+(1/x^{5})

**Find each of the following products:**

**15. (x ^{2} -3x + 7) Ã— (2x + 3)**

**16. (3x ^{2} + 5x – 9) Ã— (3x – 5)**

**17. (x ^{2} – xy + y^{2}) Ã— (x + y)**

**18. (x ^{2} + xy + y^{2}) Ã— (x – y)**

**Solution:**

15. Given (x^{2} -3x + 7) Ã— (2x + 3)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x^{2} -3x + 7) Ã— (2x + 3)

â‡’ 2x (x^{2} -3x + 7) + 3 (x^{2} -3x + 7)

â‡’ 2x^{3} â€“ 6x^{2} + 14x + 3x^{2}Â – 9x +21

â‡’ 2x^{3} – 3x^{2} +5x +21

16. Given (3x^{2} + 5x – 9) Ã— (3x – 5)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(3x^{2} + 5x – 9) Ã— (3x – 5)

â‡’ 3x (3x^{2} + 5x – 9) – 5 (3x^{2} + 5x – 9)

â‡’ 9x^{3} + 15x^{2} – 27x â€“ 15x^{2} – 25x +45

â‡’ 9x^{3} – 52x +45

17. Given (x^{2} – xy + y^{2}) Ã— (x + y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x^{2} – xy + y^{2}) Ã— (x + y)

â‡’ x (x^{2} – xy + y^{2}) + y (x^{2} + xy + y^{2})

â‡’ x^{3} – x^{2}y – y^{2}x + x^{2}y + y^{2}x + y^{3}

â‡’ (x^{3}+ y^{3})

18. Given (x^{2} + xy + y^{2}) Ã— (x – y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x^{2} + xy + y^{2}) Ã— (x – y)

â‡’ x (x^{2} + xy + y^{2}) – y (x^{2} + xy + y^{2})

â‡’ x^{3} + x^{2}y + y^{2}x – x^{2}y – y^{2}x + y^{3}

â‡’ (x^{3}– y^{3})

## Exercise 6C

**1. Divide:**

**(i) 24x ^{2}y^{3} by 3xy**

**(ii) 36xyz ^{2 }by -9xz**

**(iii) -72x ^{2}y^{2}z by -12xyz**

**(iv) -56mnp ^{2} by 7mnp**

**Solution: **

(i) Given 24x^{2}y^{3} by 3xy

â‡’24x^{2}y^{3 }/ (3xy)

On dividing monomial by a monomial we have divide same variables of each

Expressions.

On simplifying we get,

â‡’8xy^{2}

(ii) Given 36xyz^{2 }by -9xz

â‡’36xyz^{2 }/ (-9xz)

On dividing monomial by a monomial we have divide same variables of each

Expressions

On simplifying we get

â‡’-4yz

(iii) Given -72x^{2}y^{2}z by -12xyz

â‡’-72x^{2}y^{2}z / (-12xyz)

On dividing monomial by a monomial we have divide same variables of each

Expressions

On simplifying we get

â‡’6xy

(iv) Given -56mnp^{2} by 7mnp

â‡’-56mnp^{2} / (7mnp)

On dividing monomial by a monomial we have divide same variables of each

Expressions

On simplifying we get

â‡’-8p

**2. Divide:**

** (i) 5m ^{3}-30m^{2}+45m by 5m**

** (ii) 8x ^{2}y^{2}-6xy^{2}+10 x^{2}y^{3} by 2xy**

** (iii) 9x ^{2}y-6xy+12xy^{2} by -3xy**

** (iv) 12x ^{2 }+ 8x^{3}-6x^{2} by -2x^{2}**

** Solution: **

(i) Given 5m^{3}-30m^{2}+45m by 5m

â‡’-5m^{3}-30m^{2}+45m / (5m)

On dividing polynomial by a monomial we have divide every variables of polynomial

By monomial

On simplifying we get

â‡’m^{2}-6m+9

(ii) Given 8x^{2}y^{2}-6xy^{2}+10 x^{2}y^{3} by 2xy

â‡’8x^{2}y^{2}-6xy^{2}+10 x^{2}y^{3}/ (2xy)

On dividing polynomial by a monomial we have divide every variables of polynomial

By monomial

On simplifying we get

â‡’4xy-3y+5xy^{2}

(iii) Given 9x^{2}y-6xy+12xy^{2} by -3xy

â‡’9x^{2}y-6xy+12xy^{2} / (-3xy)

On dividing polynomial by a monomial we have divide every variables of polynomial

By monomial

On simplifying we get

â‡’-3x + 2 – 4y

(iv) Given 12x^{2 }+ 8x^{3}-6x^{2} by -2x^{2}

â‡’12x^{2 }+ 8x^{3}-6x^{2}/ (-2x^{2})

On dividing polynomial by a monomial we have divide every variables of polynomial

By monomial

On simplifying we get

â‡’-6x^{2} -4x + 3

**Write the quotient and remainder when we divide:**

** 3. (x ^{2}-4x+4) by (x-2)**

** 4. (x ^{2}-4) by (x+2)**

** 5. (x ^{2}+12x+35) by (x+7)**

** 6. (15x ^{2}+x-6) by (3x+2)**

** 7. (14x ^{2}-53x+45) by (7x-9)**

**Solution: **

3. Given (x^{2}-4x+4) by (x-2)

On dividing polynomial by a binomial we have divide every variables of polynomial

By binomial we get

Here quotient is x-2 and remainder is 0

4. Given (x^{2}-4) by (x+2)

On dividing polynomial by a binomial we have divide every variables of polynomial

By binomial

Here quotient is x-2 and remainder is 0

5. Given (x^{2}+12x+35) by (x+7)

On dividing polynomial by a binomial we have divide every variables of polynomial

By binomial

Here quotient is x+5 and remainder is 0

6. Given (15x^{2}+x-6) by (3x+2)

On dividing polynomial by a binomial we have divide every variables of polynomial

By binomial

Here quotient is 5x-3 and remainder is 0

7. Given (14x^{2}-53x+45) by (7x-9)

On dividing polynomial by a binomial we have divide every variables of polynomial

By binomial

Here quotient is 2x-5 and remainder is 0

## Exercise 6D

**1. Find each of the following products:**

**(i) (x + 6) (x +6) **

**(ii) (4x +5y) (4x + 5y)**

**(iii) (7a +9b) (7a +9b)**

**(iv) ((2/3) x + (4/5) y) ((2/3) x + (4/5) y)**

**(v) (x ^{2}+7) (x^{2}+7)**

**(vi) ((5/6) a ^{2}+2) ((5/6) a^{2}+2)**

**Solution: **

(i) Given that (x + 6) (x +6)

But we can write the given expression as (x + 6) (x +6) = (x + 6)^{2}

But we have (a + b)^{2}=a^{2}+2ab+b^{2}

On applying above identity in the given expression we get,

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

(x + 6)^{2}= x^{2}+12 x + 36

(ii) Given that (4x + 5y) (4x +5y)

But we can write the given expression as (4x + 5y) (4x +5y) = (4x + 5y)^{2}

But we have (a + b)^{2}=a^{2}+2ab+b^{2}

On applying above identity in the given expression we get,

(4x + 5y)^{2}= (4x)^{2}+2 (4x) (5y) + (5y)^{2}

(4x + 5y)^{2}= 16x^{2}+40 xy + 25y^{2}

(iii) Given that (7a +9b) (7a +9b)

But we can write the given expression as (7a +9b) (7a +9b) = (7a +9b)^{2}

But we have (a + b)^{2}=a^{2}+2ab+b^{2}

On applying above identity in the given expression we get,

(7a +9b)^{2}= (7a)^{2}+2 (7a) (9b) + (9b)^{2}

(7a +9b)^{2}= 49a^{2}+126 ab + 81b^{2}

(iv) Given that ((2/3) x + (4/5) y) ((2/3) x + (4/5) y)

But we can write the given expression as

((2/3) x + (4/5) y) ((2/3) x + (4/5) y)= ((2/3) x + (4/5) y)^{2}

But we have (a + b)^{2}=a^{2}+2ab+b^{2}

On applying above identity in the given expression we get,

((2/3) x + (4/5) y)^{2}= ((2/3) x)^{2}+2 ((2/3)x) ((4/5)y) + ((4/5)y)^{2}

((2/3) x + (4/5) y)^{2}= (4/9) x^{2 }+ (16/15) xy + (16/25) y^{2}

(v) Given that (x^{2}+7) (x^{2}+7)

But we can write the given expression as

(x^{2}+7) (x^{2}+7)= (x^{2}+7)^{2}

But we have (a + b)^{2}=a^{2}+2ab+b^{2}

On applying above identity in the given expression we get,

(x^{2}+7)^{2}= (x^{2})^{2}+2 ((x^{2}) (7) + (7)^{2}

(x^{2}+7)^{2}= x^{4}+14x^{2} + 49

(vi) Given that ((5/6) a^{2}+2) ((5/6) a^{2}+2)

But we can write the given expression as

((5/6) a^{2}+2) ((5/6) a^{2}+2)= ((5/6) a^{2}+2)^{2}

But we have (a + b)^{2}=a^{2}+2ab+b^{2}

On applying above identity in the given expression we get,

((5/6) a^{2}+2)^{2}= ((5/6) a^{2} )^{2}+2 ((5/6)a^{2} ) (2) + (2)^{2}

((5/6) a^{2}+2)^{2}= (25/36) a^{4}+ (10/3) a^{2}+ 4

**2. Find each of the following products:**

**(i) (x – 4) (x – 4)**

**(ii) (2x-3y) (2x-3y)**

**(iii) ((3/4) x – (5/6) y) ((3/4) x + (5/6) y)**

**(iv) (x – (3/x) ) ( x – (3/x) )**

**(v) ((1/3) x ^{2} – 9) ((1/3) x^{2} – 9)**

**(vi) ((1/2) y ^{2} â€“ (1/3) y) ((1/2) y^{2} â€“ (1/3)y) **

**Solution: **

(i) Given (x – 4) (x – 4)

But we can write the given expression as (x – 4) (x -4) = (x – 4)^{2}

But we have (a – b)^{2}=a^{2}-2ab+b^{2}

On applying above identity in the given expression we get,

(x – 4)^{2}= x^{2}-2 x (4) + 4^{2}

(x – 4)^{2}= x^{2}-8 x + 16

(ii) Given (2x-3y) (2x-3y)

But we can write the given expression as (2x â€“ 3y) (2x -3y) = (2x â€“ 3y)^{2}

But we have (a – b)^{2}=a^{2}-2ab+b^{2}

On applying above identity in the given expression we get,

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

(2x â€“ 3y)^{2}= 4x^{2}-12 xy + 9y^{2}

(iii) Given that ((3/4) x – (5/6) y) ((3/4) x + (5/6) y)

But we can write the given expression as

((3/4) x – (5/6) y) ((3/4) x + (5/6) y)= ((3/4) x – (5/6) y)^{2}

But we have (a – b)^{2}=a^{2}-2ab+b^{2}

On applying above identity in the given expression we get,

((3/4) x – (5/6) y)^{2}= ((3/4) x)^{2}-2 ((3/4) x) ((5/6) y) + ((5/6) y)^{2}

((3/4) x – (5/6) y)^{2}= (9/16) x^{2}– (15/12) xy + (25/36)y^{2}

(iv) Given that (x – (3/x)) (x – (3/x))

But we can write the given expression as

(x – (3/x)) (x – (3/x))= (x – (3/x))^{2}

But we have (a – b)^{2}=a^{2}-2ab+b^{2}

On applying above identity in the given expression we get,

(x – (3/x))^{2}= (x)^{2}-2 (x) (3/x) + (3/x)^{2}

(x – (3/x))^{2}= x^{2}– 6+ (9/x^{2})

(v) Given that ((1/3) x^{2} – 9) ((1/3) x^{2} – 9)

But we can write the given expression as

((1/3) x^{2} – 9) ((1/3) x^{2} – 9)= ((1/3) x^{2} – 9))^{ 2}

But we have (a – b)^{2}=a^{2}-2ab+b^{2}

On applying above identity in the given expression we get,

((1/3) x^{2} – 9))^{2}= ((1/3) x)^{2}-2 ((1/3) x^{2} ) (9) + (9)^{2}

((1/3) x^{2} – 9))^{2}= ((1/9) x^{4})- 6x^{2} +81

(vi) Given that ((1/2) y^{2} â€“ (1/3) y) ((1/2) y^{2} â€“ (1/3) y)

But we can write the given expression as

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

But we have (a – b)^{2}=a^{2}-2ab+b^{2}

On applying above identity in the given expression we get,

((1/2) y^{2} â€“ (1/3) y)^{ 2}= ((1/2) y^{2})^{2}Â -2 ((1/2) y^{2}) (1/3) + (1/3) y^{2}

((1/2) y^{2} â€“ (1/3) y)^{ 2}= ((1/4) y^{4}– y^{3} (1/3) + (1/9) y^{2}

3. **Expand: **

** (i) (8a+3b) ^{2}**

** (ii) (7x+2y) ^{2}**

** (iii) (5x+11) ^{2}**

** (iv) ((a/2) + (2/a)) ^{2}**

** (v) ((3x/4) + (2y/9)) ^{2}**

** (vi) (9x-10) ^{2}**

** (vii) (x ^{2}y – yz^{2})^{-2}**

** (viii) ((x/y)-((y/x)) ^{2}**

** Solution: **

(i) Given (8a+3b)^{2}

According to the identity (a + b)^{2}=a^{2}+2ab+b^{2 }we have to expand the given expression,

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

(8a+3b)^{2}= 64a^{2}+48ab+9b^{2}

(ii) Given (7x+2y)^{2 }

According to the identity (a + b)^{2}=a^{2}+2ab+b^{2 }we have to expand the given expression,

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

(7x+2y)^{2}= 49x^{2}+28xy+4y^{2}

(iii) Given (5x+11)^{2 }

According to the identity (a + b)^{2}=a^{2}+2ab+b^{2 }we have to expand the given expression,

(5x+11)^{2}= (5x)^{2}+2 (5x)(11)+(11)^{2}

(5x+11)^{2}= 25x^{2}+110x+121

(iv) Given ((a/2) + (2/a))^{2}

According to the identity (a + b)^{2}=a^{2}+2ab+b^{2 }we have to expand the given expression,

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

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

(v) Given ((3x/4) + (2y/9))^{2}

According to the identity (a + b)^{2}=a^{2}+2ab+b^{2 }we have to expand the given expression,

((3x/4) + (2y/9))^{2}= (3x/4)^{2}+2 (3x/4) (2y/9)+ (2y/9)^{2}

((3x/4) + (2y/9))^{2}= 9x^{2}/16+ (1/3) xy + (4y ^{2}/81)

(vi) Given (9x-10)^{2 }

According to the identity (a – b)^{2}=a^{2}-2ab+b^{2 }we have to expand the given expression,

(9x-10)^{2}= (9x)^{2}-2 (9x)(10)+(10)^{2}

(9x-10)^{2}= 81x^{2}-180x+100

(vii) Given (x^{2}y – yz^{2})^{-2}

According to the identity (a – b)^{2}=a^{2}-2ab+b^{2 }we have to expand the given expression,

(x^{2}y – yz^{2})^{-2}= (x^{2}y)^{2}-2 (x^{2}y)( yz^{2})+( yz^{2})^{2}

(x^{2}y – yz^{2})^{-2}= x^{4}y^{2}-2x^{2}y^{2}z^{2}+y^{2}z^{4}

(viii) Given ((x/y)-((y/x))^{2}

According to the identity (a – b)^{2}=a^{2}-2ab+b^{2 }we have to expand the given expression,

((x/y)-((y/x))^{2}= (x/y)^{2}-2 (x/y)( y/x)+(y/x)^{2}

((x/y)-((y/x))^{2}= x^{2}/y^{2}-2+y^{2}/x^{2}

**4. Find each of the following products:**

** (i) (x+3) (x-3)**

** (ii) (2x+5) (2x-5)**

** (iii) (8+x) (8-x)**

** (iv) (7x+11y) (7x-11y)**

** (v) (5x ^{2 }+ (3/4)y^{2}) (5x^{2}-(3/4)y^{2})**

** (vi) ((4x/5)-(5y/3)) ((4x/5) + (5y/3))**

** (vii) ((x + (1/x)) ((x-(1/x))**

** (viii) ((1/x) + (1/y)) ((1/x)-(1/y))**

** (ix) (2a + (3/b)) (2a – (3/b))**

** Solution: **

(i) Given (x+3) (x-3)

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

Applying the formula we get

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

(x+3) (x-3) = x^{2}-9

(ii) Given (2x+5) (2x-5)

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

Applying the formula we get

(2x+5) (2x-5) = (2x)^{2}-5^{2}

(2x+5) (2x-5)= 4x^{2}-25

(iii) Given (8+x) (8-x)

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

Applying the formula we get

(8+x) (8-x) = (8)^{2}-x^{2}

(8+x) (8-x)= 64 – x^{2 }

(iv) Given (7x+11y) (7x-11y)

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

Applying the formula we get

(7x+11y) (7x-11y) = (7x)^{2}-(11y)^{2}

(7x+11y) (7x-11y)= 49x^{2}-121y^{2}

(v) Given (5x^{2 }+ (3/4)y^{2}) (5x^{2}-(3/4)y^{2})

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

Applying the formula we get

(5x^{2 }+ (3/4) y^{2}) (5x^{2}-(3/4) y^{2})= (5x^{2})^{2}-((3/4) y^{2})^{2}

(5x^{2 }+ (3/4) y^{2}) (5x^{2}-(3/4) y^{2})= 25x^{4}-(9/16) y^{4}

(vi) Given ((4x/5)-(5y/3)) ((4x/5) + (5y/3))

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

Applying the formula we get

((4x/5)-(5y/3)) ((4x/5) + (5y/3))= (4x/5)^{2}-((5y/3)^{2}

((4x/5)-(5y/3)) ((4x/5) + (5y/3))= (16x^{2}/25)-(25y^{2}/15)

(vii) Given ((x + (1/x)) ((x-(1/x))

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

Applying the formula we get

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

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

(viii) Given ((1/x) + (1/y)) ((1/x)-(1/y))

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

Applying the formula we get

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

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

(ix) Given (2a + (3/b)) (2a – (3/b))

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

Applying the formula we get

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

(2a + (3/b)) (2a – (3/b))= 4a^{2}-9/b^{2}

**5. Using the formula for squaring a binomial, evaluate the following:**

(i) (54)^{2}

(ii) (82)^{2}

(iii) (103)^{2}

(iv) (704)^{2}

Solution:

(i) Given (54)^{2}

But we can write 54 as 50+4

And also we know that (a + b)^{2 }= a^{2}+2ab+b^{2}

By applying the above identity we get

(54)^{2 }= (50+4)^{2}= 50^{2}+2(50) (4) +4^{2}

(50+4)^{2}=2500+400+16=2916

(ii) Given (82)^{2}

But we can write 82 as 80+2

And also we know that (a + b)^{2 }= a^{2}+2ab+b^{2}

By applying the above identity we get

(82)^{2 }= (80+2)^{2}= 80^{2}+2(80) (2) +2^{2}

(80+2)^{2}=6400+320+4=6724

(iii) Given (103)^{2}

But we can write 103 as 100+3

And also we know that (a + b)^{2 }= a^{2}+2ab+b^{2}

By applying the above identity we get

(103)^{2 }= (100+3)^{2}= 100^{2}+2(100) (3) +3^{2}

(100+3)^{2}=10000+600+9=10609

(iv) Given (704)^{2}

But we can write 704 as 700+4

And also we know that (a + b)^{2 }= a^{2}+2ab+b^{2}

By applying the above identity we get

(704)^{2 }= (700+4)^{2}= 700^{2}+2(700) (4) +4^{2}

(700+4)^{2}=490000+2800+16=495616

**6. using the formula for squaring a binomial, evaluate the following:**

(i) (69)^{2}

(ii) (78)^{2}

(iii) (197)^{2}

(iv) (999)^{2}

Solution:

(i) Given (69)^{2}

But we can write 69 as 70-1

And also we know that (a – b)^{2 }= a^{2}-2ab+b^{2}

By applying the above identity we get

(69)^{2 }= (70-1)^{2}= 70^{2}-2(70) (1) +1^{2}

(70-1)^{2}=4900-140+1=4761

(ii) Given (78)^{2}

But we can write 78 as 80-2

And also we know that (a – b)^{2 }= a^{2}-2ab+b^{2}

By applying the above identity we get

(78)^{2 }= (80-2)^{2}= 80^{2}-2(80) (2) +2^{2}

(80-2)^{2}=6400-320+4=6084

(iii) Given (197)^{2}

But we can write 197 as 200-3

And also we know that (a – b)^{2 }= a^{2}-2ab+b^{2}

By applying the above identity we get

(197)^{2 }= (200-3)^{2}= 200^{2}-2(200) (3) +3^{2}

(200-3)^{2}=40000-1200+9=38809

(iv) Given (999)^{2}

But we can write 999 as 1000-1

And also we know that (a – b)^{2 }= a^{2}-2ab+b^{2}

By applying the above identity we get

(999)^{2 }= (1000-1)^{2}= 1000^{2}-2(1000) (1) +1^{2}

(1000-1)^{2}=1000000-2000+1=998001

## Exercise 6E

** Select the correct answer in each of the following:**

**1. The sum of (6a+4b-c+3), (2b-3c+4), (11b-7a+2c-1) and (2c-5a-6) is**

**(a) (4a-6b+2) (b) (-3a+14b-3c+2) (c) (-6a+17b) (d) (-6a+6b+c-4)**

**Solution: **

(c) (-6a+17b)

**Explanation:**

Given (6a+4b-c+3), (2b-3c+4), (11b-7a+2c-1) and (2c-5a-6)

To add the given expression we have arrange them column wise is given below:

** 2. (3q+7p ^{2}-2r^{3}+4) â€“ (4p^{2}-2q+7r^{3}-3) =? **

** (a)(p ^{2}+2q+5r^{3}+1) (b) (11p^{2}+q+5r^{3}+1) (c) (-3p^{2}-5q+9r^{3}-7) (d) (3p^{2}+5q-9r^{3}+7)**

** Solution:**

(d) (3p^{2}+5q-9r^{3}+7)

** Explanation: **

Given (3q+7p^{2}-2r^{3}+4) â€“ (4p^{2}-2q+7r^{3}-3)

According to the rules of subtraction of algebraic equations, we have negative sign will

Becomes positive and so we have to keep the big numerical sign.

Now arrange the variables in rows we get

(3q+7p^{2}-2r^{3}+4) â€“ (4p^{2}-2q+7r^{3}-3) = (3p^{2}+5q-9r^{3}+7)

** 3. (x+5) (x-3) =?**

** (a) x ^{2}+5x-15 (b) x^{2}-3x-15 (c) x^{2}+2x+15 (d) x^{2}+2x-15**

** Solution: **

(d) x^{2}+2x-15

Explanation:

Given (x+5) (x-3)

By solving in horizontal method we get

(x+5) (x-3)=x (x-3) + 5 (x-3)

(x+5) (x-3)= x^{2}-3x+5x-15

(x+5) (x-3)= x^{2}+2x-15

** 4. (2x+3) (3x-1) =?**

** (a) (6x ^{2}+8x-3) (b) (6x^{2}+7x-3) (c) (6x^{2}-7x-3) (d) (6x^{2}-7x+3) **

** Solution: **

(b) (6x^{2}+7x-3)

** Explanation: **

Given (2x+3) (3x-1)

By solving in horizontal method we get

(2x+3) (3x-1)= 2x (3x-1) + 3 (3x-1)

(2x+3) (3x-1)= (6x^{2}+7x-3)

** 5. (x+4) (x+4) =?**

** (a) (x ^{2}+16) (b) (x^{2}+4x+16) (c) (x^{2}+8x+16) (d) (x^{2}+16x)**

**Solution: **

(c) (x^{2}+8x+16)

** Explanation: **

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

By expanding the given expression by using (a + b)^{2 }= a^{2}+2ab+b^{2 }we get

(x+4)^{2} = x^{2}+2(x) (4)+4^{2} = x^{2}+8x+16

** 6.** **(x-6) (x-6) =?**

** (a) (x ^{2}-36) (b) (x^{2}+36) (c) (x^{2}-6x+36) (d) (x^{2}-12x+36)**

** Solution: **

(d) (x^{2}-12x+36)

** Explanation: **

Given (x-6) (x-6)=(x-6)^{2}

By expanding the given expression by using (a – b)^{2 }= a^{2}-2ab+b^{2 }we get

(x-6)^{2} = x^{2}-2(x) (6)+6^{2} = x^{2}-12x+36

## RS Aggarwal Solutions for Class 8 Maths Chapter 6 – Operations on Algebraic Expressions

Chapter 6, Operations on Algebraic Expressions, contains 5 exercises. RS Aggarwal Solutions given here contains the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.

- Addition of Algebraic Expressions
- Subtraction of Algebraic Expressions
- Multiplication of Algebraic Expressions
- Division of Algebraic Expressions
- Special Identities
- Applications of Identities

### Also, Access RS Aggarwal Solutions for Class 8 Maths Chapter 6 Operations on Algebraic Expressions Exercises

### Chapter Brief RS Aggarwal Solutions for Class 8 Maths Chapter 6 – Operations on Algebraic Expressions

The RS Aggarwal Solutions for Class 8 Maths Chapter 6 â€“ Operations on Algebraic Expressions deals with basic operations such as addition, subtraction, multiplication and division on algebraic expressions. It also deals with some special identities which help in solving problems directly and quickly. These are basic identities which we are using in higher studies.