The RD Sharma Solutions for Class 8 Maths Exercise 6.6 of Chapter 6 Algebraic Expressions and Identities are provided here in PDF. BYJU’S experts have formulated these solutions, which help students improve their conceptual knowledge, and thus obtain good marks in their exams. To excel in their final exams, we suggest students practise the RD Sharma Solutions thoroughly on a regular basis. The PDFs can be downloaded easily from the links available below. In Exercise 6.6 of Chapter 6, Algebraic Expressions and Identities, we will study problems based on the identities (An identity is an equality which is true for all values of the variable).
RD Sharma Solutions for Class 8 Maths Exercise 6.6 Chapter 6 Algebraic Expressions and Identities
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EXERCISE 6.6 PAGE NO: 6.43
1. Write the following squares of binomials as trinomials:
(i) (x + 2)2
(ii) (8a + 3b)2
(iii) (2m + 1)2
(iv) (9a + 1/6)2
(v) (x + x2/2)2
(vi) (x/4 – y/3)2
(vii) (3x – 1/3x)2
(viii) (x/y – y/x)2
(ix) (3a/2 – 5b/4)2
(x) (a2b – bc2)2
(xi) (2a/3b + 2b/3a)2
(xii) (x2 – ay)2
Solution:
(i) (x + 2)2
Let us express the given expression in trinomial
x2 + 2 (x) (2) + 22
x2 + 4x + 4
(ii) (8a + 3b)2
Let us express the given expression in trinomial
(8a)2 + 2 (8a) (3b) + (3b)2
64a2 + 48ab + 9b2
(iii) (2m + 1)2
Let us express the given expression in trinomial
(2m)2 + 2 (2m) (1) + 12
4m2 + 4m + 1
(iv) (9a + 1/6)2
Let us express the given expression in trinomial
(9a)2 + 2 (9a) (1/6) + (1/6)2
81a2 + 3a + 1/36
(v) (x + x2/2)2
Let us express the given expression in trinomial
(x)2 + 2 (x) (x2/2) + (x2/2)2
x2 + x3 + 1/4x4
(vi) (x/4 – y/3)2
Let us express the given expression in trinomial
(x/4)2 – 2 (x/4) (y/3) + (y/3)2
1/16x2 – xy/6 + 1/9y2
(vii) (3x – 1/3x)2
Let us express the given expression in trinomial
(3x)2 – 2 (3x) (1/3x) + (1/3x)2
9x2 – 2 + 1/9x2
(viii) (x/y – y/x)2
Let us express the given expression in trinomial
(x/y)2 – 2 (x/y) (y/x) + (y/x)2
x2/y2 – 2 + y2/x2
(ix) (3a/2 – 5b/4)2
Let us express the given expression in trinomial
(3a/2)2 – 2 (3a/2) (5b/4) + (5b/4)2
9/4a2 – 15/4ab + 25/16b2
(x) (a2b – bc2)2
Let us express the given expression in trinomial
(a2b)2 – 2 (a2b) (bc2) + (bc2)2
a4b2 – 2a2b2c2 + b2c4
(xi) (2a/3b + 2b/3a)2
Let us express the given expression in trinomial
(2a/3b)2 + 2 (2a/3b) (2b/3a) + (2b/3a)2
4a2/9b2 + 8/9 + 4b2/9a2
(xii) (x2 – ay)2
Let us express the given expression in trinomial
(x2)2 – 2 (x2) (ay) + (ay)2
x4 – 2x2ay + a2y2
2. Find the product of the following binomials:
(i) (2x + y) (2x + y)
(ii) (a + 2b) (a – 2b)
(iii) (a2 + bc) (a2 – bc)
(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)
(v) (2x + 3/y) (2x – 3/y)
(vi) (2a3 + b3) (2a3 – b3)
(vii) (x4 + 2/x2) (x4 – 2/x2)
(viii) (x3 + 1/x3) (x3 – 1/x3)
Solution:
(i) (2x + y) (2x + y)
Let us find the product of the given expression
2x (2x + y) + y (2x + y)
4x2 + 2xy + 2xy + y2
4x2 + 4xy + y2
(ii) (a + 2b) (a – 2b)
Let us find the product of the given expression
a (a – 2b) + 2b (a – 2b)
a2 – 2ab + 2ab – 4b2
a2 – 4b2
(iii) (a2 + bc) (a2 – bc)
Let us find the product of the given expression
a2 (a2 – bc) + bc (a2 – bc)
a4 – a2bc + bca2 – b2c2
a4 – b2c2
(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)
Let us find the product of the given expression
4x/5 (4x/5 + 3y/4) – 3y/4 (4x/5 + 3y/4)
16/25x2 + 12/20yx – 12/20xy – 9y2/16
16/25x2 – 9/16y2
(v) (2x + 3/y) (2x – 3/y)
Let us find the product of the given expression
2x (2x – 3/y) + 3/y (2x – 3/y)
4x2 – 6x/y + 6x/y – 9/y2
4x2 – 9/y2
(vi) (2a3 + b3) (2a3 – b3)
Let us find the product of the given expression
2a3 (2a3 – b3) + b3 (2a3 – b3)
4a6 – 2a3b3 + 2a3b3 – b6
4a6 – b6
(vii) (x4 + 2/x2) (x4 – 2/x2)
Let us find the product of the given expression
x4 (x4 – 2/x2) + 2/x2 (x4 – 2/x2)
x8 – 2x2 + 2x2 – 4/x4
(x8 – 4/x4)
(viii) (x3 + 1/x3) (x3 – 1/x3)
Let us find the product of the given expression
x3 (x3 – 1/x3) + 1/x3 (x3 – 1/x3)
x6 – 1 + 1 – 1/x6
x6 – 1/x6
3. Using the formula for squaring a binomial, evaluate the following:
(i) (102)2
(ii) (99)2
(iii) (1001)2
(iv) (999)2
(v) (703)2
Solution:
(i) (102)2
We can express 102 as 100 + 2
So, (102)2 = (100 + 2)2
Upon simplification, we get,
(100 + 2)2 = (100)2 + 2 (100) (2) + 22
= 10000 + 400 + 4
= 10404
(ii) (99)2
We can express 99 as 100 – 1
So, (99)2 = (100 – 1)2
Upon simplification, we get,
(100 – 1)2 = (100)2 – 2 (100) (1) + 12
= 10000 – 200 + 1
= 9801
(iii) (1001)2
We can express 1001 as 1000 + 1
So, (1001)2 = (1000 + 1)2
Upon simplification, we get,
(1000 + 1)2 = (1000)2 + 2 (1000) (1) + 12
= 1000000 + 2000 + 1
= 1002001
(iv) (999)2
We can express 999 as 1000 – 1
So, (999)2 = (1000 – 1)2
Upon simplification, we get,
(1000 – 1)2 = (1000)2 – 2 (1000) (1) + 12
= 1000000 – 2000 + 1
= 998001
(v) (703)2
We can express 700 as 700 + 3
So, (703)2 = (700 + 3)2
Upon simplification, we get,
(700 + 3)2 = (700)2 + 2 (700) (3) + 32
= 490000 + 4200 + 9
= 494209
4. Simplify the following using the formula: (a – b) (a + b) = a2 – b2:
(i) (82)2 – (18)2
(ii) (467)2 – (33)2
(iii) (79)2 – (69)2
(iv) 197 × 203
(v) 113 × 87
(vi) 95 × 105
(vii) 1.8 × 2.2
(viii) 9.8 × 10.2
Solution:
(i) (82)2 – (18)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(82)2 – (18)2 = (82 – 18) (82 + 18)
= 64 × 100
= 6400
(ii) (467)2 – (33)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(467)2 – (33)2 = (467 – 33) (467 + 33)
= (434) (500)
= 217000
(iii) (79)2 – (69)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(79)2 – (69)2 = (79 + 69) (79 – 69)
= (148) (10)
= 1480
(iv) 197 × 203
We can express 203 as 200 + 3 and 197 as 200 – 3
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
197 × 203 = (200 – 3) (200 + 3)
= (200)2 – (3)2
= 40000 – 9
= 39991
(v) 113 × 87
We can express 113 as 100 + 13 and 87 as 100 – 13
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
113 × 87 = (100 – 13) (100 + 13)
= (100)2 – (13)2
= 10000 – 169
= 9831
(vi) 95 × 105
We can express 95 as 100 – 5 and 105 as 100 + 5
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
95 × 105 = (100 – 5) (100 + 5)
= (100)2 – (5)2
= 10000 – 25
= 9975
(vii) 1.8 × 2.2
We can express 1.8 as 2 – 0.2 and 2.2 as 2 + 0.2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
1.8 × 2.2 = (2 – 0.2) ( 2 + 0.2)
= (2)2 – (0.2)2
= 4 – 0.04
= 3.96
(viii) 9.8 × 10.2
We can express 9.8 as 10 – 0.2 and 10.2 as 10 + 0.2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
9.8 × 10.2 = (10 – 0.2) (10 + 0.2)
= (10)2 – (0.2)2
= 100 – 0.04
= 99.96
5. Simplify the following using the identities:
(i) ((58)2 – (42)2)/16
(ii) 178 × 178 – 22 × 22
(iii) (198 × 198 – 102 × 102)/96
(iv) 1.73 × 1.73 – 0.27 × 0.27
(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726
Solution:
(i) ((58)2 – (42)2)/16
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
((58)2 – (42)2)/16 = ((58-42) (58+42)/16)
= ((16) (100)/16)
= 100
(ii) 178 × 178 – 22 × 22
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
178 × 178 – 22 × 22 = (178)2 – (22)2
= (178-22) (178+22)
= 200 × 156
= 31200
(iii) (198 × 198 – 102 × 102)/96
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(198 × 198 – 102 × 102)/96 = ((198)2 – (102)2)/96
= ((198-102) (198+102))/96
= (96 × 300)/96
= 300
(iv) 1.73 × 1.73 – 0.27 × 0.27
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
1.73 × 1.73 – 0.27 × 0.27 = (1.73)2 – (0.27)2
= (1.73-0.27) (1.73+0.27)
= 1.46 × 2
= 2.92
(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(8.63 × 8.63 – 1.37 × 1.37)/0.726 = ((8.63)2 – (1.37)2)/0.726
= ((8.63-1.37) (8.63+1.37))/0.726
= (7.26 × 10)/0.726
= 72.6/0.726
= 100
6. Find the value of x, if:
(i) 4x = (52)2 – (48)2
(ii) 14x = (47)2 – (33)2
(iii) 5x = (50)2 – (40)2
Solution:
(i) 4x = (52)2 – (48)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
4x = (52)2 – (48)2
4x = (52 – 48) (52 + 48)
4x = 4 × 100
4x = 400
x = 100
(ii) 14x = (47)2 – (33)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
14x = (47)2 – (33)2
14x = (47 – 33) (47 + 33)
14x = 14 × 80
x = 80
(iii) 5x = (50)2 – (40)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
5x = (50)2 – (40)2
5x = (50 – 40) (50 + 40)
5x = 10 × 90
5x = 900
x = 180
7. If x + 1/x =20, find the value of x2 + 1/ x2.
Solution:
We know that x + 1/x = 20
So when squaring both sides, we get
(x + 1/x)2 = (20)2
x2 + 2 × x × 1/x + (1/x)2 = 400
x2 + 2 + 1/x2 = 400
x2 + 1/x2 = 398
8. If x – 1/x = 3, find the values of x2 + 1/ x2 and x4 + 1/ x4.
Solution:
We know that x – 1/x = 3
So, when squaring both sides, we get
(x – 1/x)2 = (3)2
x2 – 2 × x × 1/x + (1/x)2 = 9
x2 – 2 + 1/x2 = 9
x2 + 1/x2 = 9+2
x2 + 1/x2 = 11
Now, again when we square on both sides, we get,
(x2 + 1/x2)2 = (11)2
x4 + 2 × x2 × 1/x2 + (1/x2)2 = 121
x4 + 2 + 1/x4 = 121
x4 + 1/x4 = 121-2
x4 + 1/x4 = 119
∴ x2 + 1/x2 = 11
x4 + 1/x4 = 119
9. If x2 + 1/x2 = 18, find the values of x + 1/ x and x – 1/ x.
Solution:
We know that x2 + 1/x2 = 18
When adding 2 on both sides, we get
x2 + 1/x2 + 2 = 18 + 2
x2 + 1/x2 + 2 × x × 1/x = 20
(x + 1/x)2 = 20
x + 1/x = √20
When subtracting 2 from both sides, we get
x2 + 1/x2 – 2 × x × 1/x = 18 – 2
(x – 1/x)2 = 16
x – 1/x = √16
x – 1/x = 4
10. If x + y = 4 and xy = 2, find the value of x2 + y2
Solution:
We know that x + y = 4 and xy = 2
Upon squaring on both sides of the given expression, we get
(x + y)2 = 42
x2 + y2 + 2xy = 16
x2 + y2 + 2 (2) = 16 (since xy=2)
x2 + y2 + 4 = 16
x2 + y2 = 16 – 4
x2 + y2 =12
11. If x – y = 7 and xy = 9, find the value of x2+y2
Solution:
We know that x – y = 7 and xy = 9
Upon squaring on both sides of the given expression, we get
(x – y)2 = 72
x2 + y2 – 2xy = 49
x2 + y2 – 2 (9) = 49 (since xy=9)
x2 + y2 – 18 = 49
x2 + y2 = 49 + 18
x2 + y2 =67
12. If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2
Solution:
We know that 3x + 5y = 11 and xy = 2
Upon squaring on both sides of the given expression, we get
(3x + 5y)2 = 112
(3x)2 + (5y)2 + 2(3x)(5y) = 121
9x2 + 25y2 + 2 (15xy) = 121 (since xy=2)
9x2 + 25y2 + 2(15(2)) = 121
9x2 + 25y2 + 60 = 121
9x2 + 25y2 = 121-60
9x2 + 25y2 = 61
13. Find the values of the following expressions:
(i) 16x2 + 24x + 9 when x = 7/4
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = 4/3
(iii) 81x2 + 16y2 – 72xy when x = 2/3 and y = ¾
Solution:
(i) 16x2 + 24x + 9 when x = 7/4
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(4x)2 + 2 (4x) (3) + 32
(4x + 3)2
Evaluating when x = 7/4
[4 (7/4) + 3]2(7 + 3)2
100
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = 4/3
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(8x)2 + 2 (8x) (9y) + (9y)2 (8x + 9y)
Evaluating when x = 11 and y = 4/3
[8 (11) + 9 (4/3)]2(88 + 12)2
(100)2
10000
(iii) 81x2 + 16y2 – 72xy when x = 2/3 and y = ¾
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(9x)2 + (4y)2 – 2 (9x) (4y)
(9x – 4y)2
Putting x = 2/3 and y = 3/4
[9 (2/3) – 4 (3/4)]2(6 – 3)2
32
9
14. If x + 1/x = 9 find the value of x4 + 1/ x4.
Solution:
We know that x + 1/x = 9
So, when squaring both sides, we get
(x + 1/x)2 = (9)2
x2 + 2 × x × 1/x + (1/x)2 = 81
x2 + 2 + 1/x2 = 81
x2 + 1/x2 = 81 – 2
x2 + 1/x2 = 79
Now, again when we square on both sides, we get,
(x2 + 1/x2)2 = (79)2
x4 + 2 × x2 × 1/x2 + (1/x2)2 = 6241
x4 + 2 + 1/x4 = 6241
x4 + 1/x4 = 6241- 2
x4 + 1/x4 = 6239
∴ x4 – 1/x4 = 6239
15. If x + 1/x = 12 find the value of x – 1/x.
Solution:
We know that x + 1/x = 12
So, when squaring both sides, we get
(x + 1/x)2 = (12)2
x2 + 2 × x × 1/x + (1/x)2 = 144
x2 + 2 + 1/x2 = 144
x2 + 1/x2 = 144 – 2
x2 + 1/x2 = 142
When subtracting 2 from both sides, we get
x2 + 1/x2 – 2 × x × 1/x = 142 – 2
(x – 1/x)2 = 140
x – 1/x = √140
16. If 2x + 3y = 14 and 2x – 3y = 2, find value of xy. [Hint: Use (2x+3y) 2 – (2x-3y) 2 = 24xy]
Solution:
We know that the given equations are
2x + 3y = 14… equation (1)
2x – 3y = 2… equation (2)
Now, let us square both the equations and subtract equation (2) from equation (1), we get,
(2x + 3y) 2 – (2x – 3y) 2 = (14)2 – (2)2
4x2 + 9y2 + 12xy – 4x2 – 9y2 + 12xy = 196 – 4
24 xy = 192
xy = 8
∴ the value of xy is 8.
17. If x2 + y2 = 29 and xy = 2, find the value of
(i) x + y
(ii) x – y
(iii) x4 + y4
Solution:
(i) x + y
We know that
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x + y) 2 – 2 (2) = 29
(x + y) 2 = 29 + 4
x + y = ± √33
(ii) x – y
We know that
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x – y)2 + 2 (2) = 29
(x – y)2 + 4 = 29
(x – y)2 = 25
(x – y) = ± 5
(iii) x4 + y4
We know that
x2 + y2 = 29
Squaring both sides, we get
(x2 + y2)2 = (29)2
x4 + y4 + 2x2y2 = 841
x4 + y4 + 2 (2)2 = 841
x4 + y4 = 841 – 8
x4 + y4 = 833
18. What must be added to each of the following expressions to make it a whole square?
(i) 4x2 – 12x + 7
(ii) 4x2 – 20x + 20
Solution:
(i) 4x2 – 12x + 7
(2x)2 – 2 (2x) (3) + 32 – 32 + 7
(2x – 3)2 – 9 + 7
(2x – 3)2 – 2
∴ 2 must be added to the expression to make it a whole square.
(ii) 4x2 – 20x + 20
(2x)2 – 2 (2x) (5) + 52 – 52 + 20
(2x – 5)2 – 25 + 20
(2x – 5)2 – 5
∴ 5 must be added to the expression to make it a whole square.
19. Simplify:
(i) (x – y) (x + y) (x2 + y2) (x4 + y4)
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
(iii) (7m – 8n) 2 + (7m + 8n) 2
(iv) (2.5p – 1.5q) 2 – (1.5p – 2.5q) 2
(v) (m2 – n2m) 2 + 2m3n2
Solution:
(i) (x – y) (x + y) (x2 + y2) (x4 + y4)
By grouping the values
(x2 – y2) (x2 + y2) (x4 + y4)
[(x2)2 – (y2)2] (x4 + y4)(x4 – y4) (x4 – y4)
[(x4)2 – (y4)2]x8 – y8
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
Let us simplify the expression by grouping
[(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)(4x2 – 1) (4x2 + 1) (16x4 + 1) 1
[(4x2)2 – (1)2] (16x4 + 1) 1(16x4 – 1) (16x4 + 1) 1
[(16x4)2 – (1)2] 1256x8 – 1
(iii) (7m – 8n)2 + (7m + 8n)2
Upon expansion
(7m)2 + (8n)2 – 2(7m)(8n) + (7m)2 + (8n)2 + 2(7m)(8n)
(7m)2 + (8n)2 – 112mn + (7m)2 + (8n)2 + 112mn
49m2 + 64n2 + 49m2 + 64n2
By grouping the similar expression, we get,
98m2 + 64n2 + 64n2
98m2 + 128n2
(iv) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
Upon expansion
(2.5p)2 + (1.5q)2 – 2 (2.5p) (1.5q) – (1.5p)2 – (2.5q)2 + 2 (1.5p) (2.5q)
6.25p2 + 2.25q2 – 2.25p2 – 6.25q2
By grouping the similar expression, we get,
4p2 – 6.25q2 + 2.25q2
4p2 – 4q2
4 (p2 – q2)
(v) (m2 – n2m)2 + 2m3n2
Upon expansion using (a + b) 2 formula
(m2)2 – 2 (m2) (n2) (m) + (n2m) 2 + 2m3n2
m4 – 2m3n2 + (n2m)2 + 2m3n2
m4+ n4m2 – 2m3n2 + 2m3n2
m4+ m2n4
20. Show that:
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9a – 5b)2 + 180ab = (9a + 5b)2
(iii) (4m/3 – 3n/4)2 + 2mn = 16m2/9 + 9n2/16
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:
(i) (3x + 7)2 – 84x = (3x – 7)2
Let us consider LHS (3x + 7)2 – 84x
By using the formula (a + b)2 = a2 + b2 + 2ab
(3x)2 + (7)2 + 2 (3x) (7) – 84x
(3x)2 + (7)2 + 42x – 84x
(3x)2 + (7)2 – 42x
(3x)2 + (7)2 – 2 (3x) (7)
(3x – 7)2 = R.H.S
Hence, proved
(ii) (9a – 5b)2 + 180ab = (9a + 5b)2
Let us consider LHS (9a – 5b)2 + 180ab
By using the formula (a + b)2 = a2 + b2 + 2ab
(9a)2 + (5b)2 – 2 (9a) (5b) + 180ab
(9a)2 6 (5b)2 – 90ab + 180ab
(9a)2 + (5b)2 + 9ab
(9a)2 + (5b)2 + 2 (9a) (5b)
(9a + 5b)2 = R.H.S
Hence, proved
(iii) (4m/3 – 3n/4)2 + 2mn = 16m2/9 + 9n2/16
Let us consider LHS (4m/3 – 3n/4)2 + 2mn
(4m/3)2 + (3n/4)2 – 2mn + 2mn
(4m/3)2 + (3n/4)2
16/9m2 + 9/16n2 = R.H.S
Hence, proved
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
Let us consider LHS (4pq + 3q)2 – (4pq – 3q)2
(4pq)2 + (3q)2 + 2 (4pq) (3q) – (4pq)2 – (3q)2 + 2(4pq)(3q)
24pq2 + 24pq2
48pq2 = RHS
Hence, proved
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Let us consider LHS (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
By using the identity (a – b) (a + b) = a2 – b2
We get,
(a2 – b2) + (b2 – c2) + (c2 – a2)
a2 – b2 + b2 – c2 + c2 – a2
0 = R.H.S
Hence, proved
It is very helpful