Class 8 SELINA Solutions Maths Chapter 12: Identities
Identities Exercise Ex. 12(A)
Solution 1(i)
Correct option: (c) 2x2 + 8y2
Solution 1(ii)
Correct option: (c) 0
Solution 1(iii)
Correct option: (c) –48xy
Solution 1(iv)
Correct option: (c) 0.36
(0.8)2 – 0.32 + (0.2)2
= (0.8)2 – 2 × 0.8 × 0.2 + (0.2)2
= (0.8 – 0.2)2
= (0.6)2
= 0.36
Solution 1(v)
Correct option: (b) a2 + b2 – c2 – 2ab
(a – b – c)(a – b + c)
= (a – b)2 – (c)2
= a2 + b2 – 2ab – c2
Solution 2(i)
(a – 8)(a + 2) = a2 + (–8 + 2)a + (–8)(2)
= a2 – 6a – 16
Solution 2(ii)
(b – 3)(b – 5) = b2 – (3 + 5)b + (3)(5)
= b2 – 8b + 15
Solution 2(iii)
(3x – 2y)(2x + y) = (3x)(2x) + [3x × y + (–2y) × 2x] + (–2y)(y)
= 6x2 + (3xy – 4xy) – 2y2
= 6x2 – xy – 2y2
Solution 2(iv)
(5a + 16)(3a – 7) = (5a)(3a) + [5a × (–7) + 16 × 3a] + (16)(–7)
= 15a2 + (–35a + 48a) – 112
= 15a2 + 13a – 112
Solution 2(v)
(8 – b)(3 + b) = (8)(3) + [8 × b + (–b) × 3] + (–b)(b)
= 24 + (8b – 3b) – b2
= 24 + 5b – b2
Solution 3(i)
(2a + 3)(2a – 3) = (2a)2 – (3)2 = 4a2 – 9
Solution 3(ii)
(xy + 4)(xy – 4) = (xy)2 – (4)2 = x2y2 – 16
Solution 3(iii)
(ab + x2)(ab – x2) = (ab)2 – (x2)2 = a2b2 – x4
Solution 3(iv)
(3x2 + 5y2)(3x2 – 5y2) = (3x2)2 – (5y2)2 = 9x4 – 25y4
Solution 3(v)
Solution 3(vi)
Solution 3(vii)
(0.5 – 2a)(0.5 + 2a) = (0.5)2 – (2a)2 = 0.25 – 4a2
Solution 3(viii)
Solution 4(i)
(a + b)(a – b)(a2 + b2) = (a2 – b2)(a2 + b2)
= (a2)2 – (b2)2
= a4 – b4
Solution 4(ii)
(3 – 2x)(3 + 2x)(9 + 4x2) = [(3)2 – (2x)2](9 + 4x2)
= (9 – 4x2)(9 + 4x2)
= (9)2 – (4x2)2
= 81 – 16x4
Solution 4(iii)
(3x – 4y)(3x + 4y)(9x2 + 16y2) = [(3x)2 – (4y)2](9x2 + 16y2)
= (9x2 – 16y2)(9x2 + 16y2)
= (9x2)2 – (16y2)2
= 81x4 – 256y4
Solution 5(i)
21 × 19 = (20 + 1)(20 – 1)
= (20)2 – (1)2
= 400 – 1
= 399
Solution 5(ii)
33 × 27 = (30 + 3)(30 – 3)
= (30)2 – (3)2
= 900 – 9
= 891
Solution 5(iii)
103 × 97 = (100 + 3)(100 – 3)
= (100)2 – (3)2
= 10000 – 9
= 9991
Solution 5(iv)
9.8 × 10.2 = (10 – 0.2)(10 + 0.2)
= (10)2 – (0.2)2
= 100 – 0.04
= 99.96
Solution 5(v)
7.7 × 8.3 = (8 – 0.3)(8 + 0.3)
= (8)2 – (0.3)2
= 64 – 0.09
= 63.91
Solution 6(i)
(6 – xy)(6 + xy) = (6)2 – (xy)2 = 36 – x2y2
Solution 6(ii)
Solution 6(iii)
Solution 6(iv)
Solution 6(v)
(2a + 3)(2a – 3)(4a2 + 9) = [(2a)2 – (3)2](4a2 + 9)
= (4a2 – 9)(4a2 + 9)
= (4a2)2 – (9)2
= 16a4 – 81
Solution 6(vi)
(a + bc)(a – bc)(a2 + b2c2) = [(a)2 – (bc)2](a2 + b2c2)
= (a2 – b2c2)(a2 + b2c2)
= (a2)2 – (b2c2)2
= a4 – b4c4
Solution 7(i)
Solution 7(ii)
Solution 7(iii)
(a + b – c)2
= a2 + b2 + c2 + 2ab – 2bc – 2ca
= a2 + b2 + c2 + 2(ab – bc – ca)
Solution 7(iv)
(a – b + c)2
= a2 + b2 + c2 – 2ab – 2bc + 2ca
= a2 + b2 + c2 + 2(ca – ab – bc)
Solution 7(v)
Solution 8(i)
Solution 8(ii)
Solution 8(iii)
(x – 2y + 1)2
= (x)2 + (–2y)2 + (1)2 + 2(x)(–2y) + 2(–2y)(1) + 2(x)(1)
= x2 + 4y2 + 1 – 4xy – 4y + 2x
Solution 8(iv)
(3a – 2b – 5c)2
= (3a)2 + (–2b)2 + (–5c)2 + 2(3a)(–2b) + 2(–2b)(–5c) + 2(3a)(–5c)
= 9a2 + 4b2 + 25c2 – 12ab + 20bc – 30ac
Solution 8(v)
Solution 8(vi)
Solution 8(vii)
(2x – 3y + z)2
= (2x)2 + (–3y)2 + (z)2 + 2(2x)(–3y) + 2(–3y)(z) + 2(2x)(z)
= 4x2 + 9y2 + z2 – 12xy – 6yz + 4xz
Solution 8(viii)
Solution 9(i)
(208)2
= (200 + 8)2
= (200)2 + 2(200)(8) + (8)2
= 40000 + 3200 + 64
= 43264
Solution 9(ii)
(92)2 = (90 + 2)2
= (90)2 + 2(90)(2) + (2)2
= 8100 + 360 + 4
= 8464
Solution 9(iii)
(9.4)2 = (10 – 0.6)2
= (10)2 – 2(10)(0.6) + (0.6)2
= 100 – 12 + 0.36
= 88.36
Solution 9(iv)
(20.7)2 = (20 + 0.7)2
= (20)2 + 2(20)(0.7) + (0.7)2
= 400 + 28 + 0.49
= 428.49
Solution 10(i)
(2a + b)3
= (2a)3 + 3(2a)2(b) + 3(2a)(b)2 + (b)3
= (2a)3 + 3(4a2)(b) + 3(2a)(b2) + (b)3
= 8a3 + 12a2b + 6ab2 + b3
Solution 10(ii)
(a – 2b)3
= (a)3 – 3(a)2(2b) + 3(a)(2b)2 – (2b)3
= a3 – 3(a2)(2b) + 3(a)(4b2) – 8b3
= a3 – 6a2b + 12ab2 – 8b3
Solution 10(iii)
(3x – 2y)3
= (3x)3 – 3(3x)2(2y) + 3(3x)(2y)2 – (2y)3
= 27x3 – 3(9x2)(2y) + 3(3x)(4y2) – 8y3
= 27x3 – 54x2y + 36xy2 – 8y3
Solution 10(iv)
(x + 5y)3
= (x)3 + 3(x)2(5y) + 3(x)(5y)2 + (5y)3
= (x)3 + 3(x2)(5y) + 3(x)(25y2) + (125y3)
= x3 + 15x2y + 75xy2 + 125y3
Solution 10(v)
Solution 10(vi)
Solution 11(i)
(a + 2)3
= a3 + 3(a)2(2) + 3(a)(2)2 + (2)3
= a3 + 6a2 + 12a + 8
Solution 11(ii)
(2a – 1)3
= (2a)3 – 3(2a)2(1) + 3(2a)(1)3 – (1)3
= 8a3 – 12a2 + 6a – 1
Solution 11(iii)
(2a + 3b)3
= (2a)3 + 3(2a)2(3b) + 3(2a)(3b)2 + (3b)3
= 8a3 + 3(4a2)(3b) + 3(2a)(9b2) + 27b3
= 8a3 + 36a2b + 54ab2 + 27b3
Solution 11(iv)
(3b – 2a)3
= (3b)3 – 3(3b)2(2a) + 3(3b)(2a)2 – (2a)3
= 27b3 – 3(9b2)(2a) + 3(3b)(4a2) – 8a3
= 27b3 – 54b2a +36ba2 – 8a3
Solution 11(v)
Solution 11(vi)
Identities Exercise Ex. 12(B)
Solution 1(i)
Correct option: (c) 11
Solution 1(ii)
Correct option: (a) 29
(a + b) = 7
⇒ (a + b)2 = 49
⇒ a2 + b2 + 2ab = 49
⇒ a2 + b2 + 2(10) = 49
⇒ a2 + b2 = 29
Solution 1(iii)
Correct option: (a) 100
Solution 1(iv)
Correct option: (b) 2
(a – b)2 = (a + b)2 – 4ab
⇒ (1)2 = (3)2 – 4ab
⇒ 1 = 9 – 4ab
⇒ 4ab = 8
⇒ ab = 2
Solution 1(v)
Correct option: (a) 0
Solution 2
(a + b) = 5
⇒ (a + b)2 = 25
⇒ a2 + b2 + 2ab = 25
⇒ a2 + b2 + 2(6) = 25
⇒ a2 + b2 + 12 = 25
⇒ a2 + b2 = 13
Solution 3
(a – b) = 6
⇒ (a – b)2 = 36
⇒ a2 + b2 – 2ab = 36
⇒ a2 + b2 – 2(16) = 36
⇒ a2 + b2 – 32 = 36
⇒ a2 + b2 = 68
Solution 4(i)
(a + b)2 = a2 + b2 + 2ab
= 29 + 2(10)
= 29 + 20
= 49
⇒ a + b = ±7
Solution 4(ii)
(a – b)2 = a2 + b2 – 2ab
= 29 – 2(10)
= 29 – 20
= 9
⇒ a – b = ±3
Solution 5(i)
(a – b)2 = a2 + b2 – 2ab
= 10 – 2(3)
= 10 – 6
= 4
⇒ a – b = ±2
Solution 5(ii)
(a + b)2 = a2 + b2 + 2ab
= 10 + 2(3)
= 10 + 6
= 16
⇒ a + b = ±4
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
(a + b + c) = 10
⇒ (a + b + c)2 = 100
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 100
⇒ 38 + 2(ab + bc + ca) = 100
⇒ 2(ab + bc + ca) = 62
⇒ (ab + bc + ca) = 31
Solution 11
(a + b + c) = 9
⇒ (a + b + c)2 = 81
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81
⇒ a2 + b2 + c2 + 2(24) = 81
⇒ a2 + b2 + c2 + 48 = 81
⇒ a2 + b2 + c2 = 33
Solution 12
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
= 83 + 2(71)
= 83 + 142
= 225
⇒ (a + b + c) = ±15
Solution 13
(a + b) = 6
⇒ (a + b)3 = 216
⇒ a3 + b3 + 3ab(a + b) = 216
⇒ a3 + b3 + 3(8)(6) = 216
⇒ a3 + b3 + 144 = 216
⇒ a3 + b3 = 72
Solution 14
(a – b) = 3
⇒ (a – b)3 = 27
⇒ a3 – b3 – 3ab(a – b) = 27
⇒ a3 – b3 – 3(10)(3) = 27
⇒ a3 – b3 – 90 = 27
⇒ a3 – b3 = 117
Solution 15
Solution 16
Solution 17(i)
Solution 17(ii)
Solution 18(i)
Solution 18(ii)
Solution 19(i)
Let the numbers be x and y.
Then, x2 + y2 = 13 and xy = 6
Now, (x + y)2 = x2 + y2 + 2xy
= 13 + 2(6)
= 13 + 12
= 25
⇒ (x + y) = ±5
Solution 19(ii)
Let the numbers be x and y.
Then, x2 + y2 = 13 and xy = 6
Now, (x – y)2 = x2 + y2 – 2xy
= 13 – 2(6)
= 13 – 12
= 1
⇒ (x – y) = ±1
Identities Exercise TEST YOURSELF
Solution 1(i)
Correct option: (a) 4
Solution 1(ii)
Correct option: (c) x8 – y8
(x + y)(x – y)(x2 + y2)(x4 + y4)
= (x2 – y2) (x2 + y2)(x4 + y4)
= (x4 – y4)(x4 + y4)
= (x8 – y8)
Solution 1(iii)
Correct option: (d) 9996
102 × 98 = (100 + 2)(100 – 2)
= (100)2 – (2)2
= 10000 – 4
= 9996
Solution 1(iv)
Correct option: (a) 10x + 5
(x + 3)(x + 3) – (x – 2)(x – 2)
= (x + 3)2 – (x – 2)2
= (x + 3 + x – 2)(x + 3 – x + 2)
= (2x + 1)(5)
= 10x + 5
Solution 1(v)
Correct option: (c) 55
5a = 302 – 252
= (30 – 25)(30 + 25)
= 5 × 55
= 275
⇒ a = 55
Solution 2(i)
Solution 2(ii)
(2a + 0.5)(7a – 0.3)
= (2a)(7a) + [2a × (–0.3) + (0.5) × 7a] + (0.5)(–0.3)
= 14a2 + (–0.6a + 3.5a) – 0.15
= 14a2 + 2.9a – 0.15
Solution 2(iii)
(9 – y)(7 + y)
= (9)(7) + [9 × y + (–y) × 7] + (–y)(y)
= 63 + (9y – 7y) – y2
= 63 + 2y – y2
Solution 2(iv)
(2 – z)(15 – z)
= (2)(15) + [2 × (–z) + (–z) × 15] + (–z)( –z)
= 30 + [–2z – 15z] + z2
= 30 – 17z + z2
Solution 2(v)
(a2 + 5)(a2 – 3)
= (a2)(a2) + [a2 × (–3) + (5) × a2] + (5)(–3)
= a4 + (–3a2 + 5a2) – 15
= a4 + 2a2 – 15
Solution 2(vi)
(4 – ab)(8 + ab)
= (4)(8) + [4 × ab + (–ab) × 8] + (–ab)(ab)
= 32 + (4ab – 8ab) – a2b2
= 32 – 4ab – a2b2
Solution 2(vii)
(5xy – 7)(7xy + 9)
= (5xy)(7xy) + [5xy × 9 + (–7) × 7xy] + (–7)(9)
= 35x2y2 + (45xy – 49xy) – 63
= 35x2y2 – 4xy – 63
Solution 2(viii)
(3a2 – 4b2)(8a2 – 3b2)
= (3a2)(8a2) + [3a2 × (–3b2) + (–4b2) × 8a2] + (–4b2)( –3b2)
= 24a4 + (–9a2b2 – 32a2b2) + 12b4
= 24a4 – 41a2b2 + 12b4
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 3(iv)
Solution 3(v)
Solution 3(vi)
(607)2 = (600 + 7)2
= (600)2 + 2(600)(7) + (7)2
= 360000 + 8400 + 49
= 368449
Solution 3(vii)
(391)2 = (400 – 9)2
= (400)2 – 2(400)(9) + (9)2
= 160000 – 7200 + 81
= 152881
Solution 3(viii)
(9.7)2 = (10 – 0.3)2
= (10)2 – 2(10)(0.3) + (0.3)2
= 100 – 6 + 0.09
= 94.09
Solution 4(i)
Solution 4(ii)
Solution 5(i)
Solution 5(ii)
Solution 5(iii)
Solution 6(i)
(a – b)2 = a2 + b2 – 2ab
= 41 – 2(4)
= 41 – 8
= 33
Solution 6(ii)
(a + b)2 = a2 + b2 + 2ab
= 41 + 2(4)
= 41 + 8
= 49
Solution 7(i)
Solution 7(ii)
Solution 8(i)
Solution 8(ii)
Solution 9(i)
(3x – 4y + 5z)2
= (3x)2 + (–4y)2 + (5z)2 + 2(3x)(–4y) + 2(–4y)(5z) + 2(3x)(5z)
= 9x2 + 16y2 + 25z2 – 24xy – 40yz + 30zx
Solution 9(ii)
(2a – 5b – 4c)2
= (2a)2 + (–5b)2 + (–4c)2 + 2(2a)(–5b) + 2(–5b)(–4c) + 2(2a)(–4c)
= 4a2 + 25b2 + 16c2 – 20ab + 40bc – 16ca
Solution 9(iii)
(5x + 3y)3
= (5x)3 + (3y)3 + 3(5x)(3y)(5x + 3y)
= 125x3 + 27y3 + 45xy(5x + 3y)
= 125x3 + 27y3 + 225x2y + 135xy2
Solution 9(iv)
(6a – 7b)3
= (6a)3 + (–7b)3 + 3(6a)(–7b)(6a – 7b)
= 216a3 – 343b3 – 126ab(6a – 7b)
= 216a3 – 343b3 – 756a2b + 882ab2
Solution 10
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(9)2 = a2 + b2 + c2 + 2(15)
81 = a2 + b2 + c2 + 30
a2 + b2 + c2 = 81 – 30 = 51
Solution 11
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(11)2 = 81 + 2(ab + bc + ca)
121 = 81 + 2(ab + bc + ca)
2(ab + bc + ca) = 40
ab + bc + ca = 20
Solution 12
(3x – 4y)3 =(3x)3 – (4y)3 – 3(3x)(4y)(3x – 4y)
53 = 27x3 – 64y3 – 36xy(5)
125 = 27x3 – 64y3 – 36(3)(5)
125 = 27x3 – 64y3 – 540
27x3 – 64y3 = 665
Solution 13
(a + b)3 = a3 + b3 + 3(ab)(a + b)
83 = a3 + b3 + 3(15)(8)
512 = a3 + b3 + 360
a3 + b3 = 152
Solution 14
(3x + 2y)3 = (3x)3 + (2y)3 + 3(3x)(2y)(3x + 2y)
93 = 27x3 + 8y3 + 18xy(9)
729 = 27x3 + 8y3 + 18(3)(9)
729 = 27x3 + 8y3 + 486
27x3 + 8y3 = 243
Solution 15
(5x – 4y)3 = (5x)3 – (4y)3 – 3(5x)(4y)(5x – 4y)
73 = 125x3 – 64y3 – 60xy(7)
343 = 125x3 – 64y3 – 420(8)
343 = 125x3 – 64y3 – 3360
125x3 – 64y3 = 3703
Solution 16
Let the numbers be x and y.
Then, x – y = 5 and xy = 14
(x – y)3 = (x)3 – (y)3 – 3xy(x – y)
53 = x3 – y3 – 3(14)(5)
125 = x3 – y3 – 210
x3 – y3 = 335