Algebraic Identities ICSE Class-8th Concise Selina Mathematics Solutions Chapter-12 . We provide step by step Solutions of Exercise / lesson-12 Algebraic Identities for ICSE Class-8 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-12 A, Exe-12 B, Exe-12 C and Exe-12 D to develop skill and confidence . Visit official Website CISCE for detail information about ICSE Board Class-8.
Algebraic Identities ICSE Class-8th Concise Selina Mathematics Solutions Chapter-12
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ICSE Class-8 Algebraic Identities Exe-12 A,
Question 1 :-
Use direct method to evaluate the following products :
(i) (x + 8)(x + 3)
(ii) (y + 5)(y – 3)
(iii) (a – 8)(a + 2)
(iv) (b – 3)(b – 5)
(v) (3x – 2y)(2x + y)
(vi) (5a + 16)(3a – 7)
(vii) (8 – b) (3 + b)
Answer :-
(i)
(x + 8) (x + 3)
= (x × x) + (x × 3) + (8 × x) + (8 × 3)
= x2 + 3x + 8x + 24
= x2 + 11x + 24
(ii)
(y + 5) (y – 3) = (y × y) + (y × −3) + (5 × y) + (5 × −3)
= y2 + (−3y) + (5y) − 15
= y2 − 3y + 5y − 15
= y2 + 2y − 15
(iii)
(a – 8) (a + 2) = (a × a) + (a × 2) + (−8)×a + (−8)(2)
= a2 + 2a − 8a − 16
= a2 − 6a − 16
(iv)
(b – 3) (b – 5) = (b × b) + (b × −5) + (−3) × b + (−3) (−5)
= b2 − 5b − 3b + 15
= b2 − 8b + 15
(v)
(3x – 2y) (2x + y) = (3x × 2x) + (3x × y) + (−2y × 2x) + (−2y × y)
= 6x2 + 3xy − 4xy − 2y2
= 6x2 − xy − 2y2
(vi)
(5a + 16) (3a – 7) = (5a × 3a) + (5a × −7) + (16 × 3a) + 16 −7
= 15a2 + (−35a) + 48a + (−112)
= 15a2 − 35a + 48a − 112
= 15a2 + 13a − 112
(vii)
(8 – b) (3 + b) = (8 × 3) + (8 × b) + (−b × 3) + (−b × b)
= 24 + 8b − 3b − b2
= 24 + 5b − b2
Question 2 :-
Use direct method to evaluate :
Answer :-
(i)
(a+b) (a−b) = a2 − b2
(x+1) (x−1) = (x)2 − (1)2
= x2 − 1
(ii)
(a+b) (a−b) = a2 − b2
(2+a) (2−a) = (2)2 − (a)2
= 4 − a2
(iii)
(a+b) (a−b) = a2 − b2
(3b−1) (3b+1)
= (3b)2 − (1)2
= 9b2 − 1
(iv)
(a+b) (a−b) = a2 − b2
(4+5x) (4−5x)
= (4)2 − (5x)2
= 16 − 25x2
(v)
(a+b) (a−b) = a2 − b2
(2a+3) (2a−3) = (2a)2 − (3)2
= 4a2 − 9
(vi)
(a+b) (a−b) = a2 − b2
(xy+4) (xy−4) = (xy)2 − (4)2
= x2y2 − 16
(vii)
(a+b) (a−b) = a2 − b2
(ab+x2) (ab−x2) = (ab)2 − (x2)2
= a2b2 − x4
(viii)
(a+b) (a−b) = a2 − b2
(3x2+5y2) (3x2−5y2) = (3x2)2 − (5y2)2
= 9x4 − 25y4
(ix)
(a+b) (a−b) = a2 − b2
(x)
(a+b) (a−b) = a2 − b2
(xi)
(a+b) (a−b) = a2 − b2
(0.5a−2a) (0.5+2a)
= (0.5)2 − (2a)2
= 0.25 − 4a2
(xii)
(a+b) (a−b) = a2 − b2
Question 3 :-
Evaluate :
Answer :-
(i)
(a+1) (a-1) (a2+1)
= [(a)2−(1)2] (a2+1)
= (a2−1) (a2+1)
= (a2)2 − (1)2
= a4 − 1
(ii)
(a+b) (a−b) (a2+b2)
= (a2−b2) (a2+b2)
= (a2)2 − (b2)2
= a4 − b4
(iii)
(2a−b) (2a+b) (4a2+b2)
= [(2a)2−(b)2] (4a2+b2)
= (4a2−b2) (4a2+b2)
= (4a2)2 − (b2)2
= 16a4 − b4
(iv)
(3−2x) (3+2x) (9+4x2)
= [{3}2−(2x)2] (9+4x2)
= (9−4x2) (9+4x2)
= (9)2 − (4x2)2
= 81 − 16x4
(v)
(3−2x) (3+2x) (9+4x2)
= [{3}2−(2x)2] (9+4x2)
= (9−4x2) (9+4x2)
= (9)2 − (4x2)2
= 81 − 16x4
Question 4 :-
Use the product (a + b) (a – b) = a2 – b2 to evaluate:
(i) 21 × 19
(ii) 33 × 27
(iii) 103 × 97
(iv) 9.8 × 10.2
(v) 7.7 × 8.3
(vi) 4.6 × 5.4
Answer :-
(i)
21 x 19 = (20+1) (20−1)
= (20)2 − (1)2
= 400 − 1
= 399
(ii)
33 x 27 = (30+3) (30−3)
= (30)2 − (3)2
= 900 − 9
= 891
(iii)
103 x 97 = (100+3) (100−3)
= (100)2 − (3)2
= 10000 − 9
= 9991
(iv)
9.8 x 10.2 = (10−0.2) (10+0.2)
= (10)2 − (0.2)2
= 100 − 0.04
= 99.96
(v)
7.7 x 8.3 = (8−0.3) (8+0.3)
= (8)2 − (0.3)2
= 64 − 0.09
= 63.91
(vi)
4.6 x 5.4 = (5−0.4) (5+0.4)
= (5)2 − (0.4)2
= 25 − 0.16
= 24.84
Question 5 :-
Evaluate :
(i) (6 – xy) (6 + xy)
Answer :-
(i)
(6 – xy) (6 + xy) = 6(6+xy) − xy(6+xy)
= 36 + 6xy − 6xy + (xy)2
= 36 − x2 − y2
(ii) 
(iii) 
(iv) 
(v)
(2a +3) (2a − 3) (4a2 + 9)
= [(2a)2 − (3)2] (4a2 + 9) ……….[(a+b) (a−b) = a2 − b2]
= (4a2 − 9) (4a2 + 9)
= (4a2)2 − (9)2 ………[(a+b) (a−b) = a2 − b2]
= 16a4 − 81
(vi)
(a + bc) (a − bc) (a2 + b2c2)
= [(a)2 − (bc)2] (a2 + b2c2) ………..[(a+b) (a−b) = a2 − b2]
= (a2 − b2c2) (a2 + b2c2) ……….[ ∵ (a+b) (c−b) = a2 − b2]
= a4 − b4c4
(vii)
(5x + 8y) (3x + 5y)
= 5x (3x + 5y) + 8y (3x+ 5y)
= 15x2 + 25xy + 24xy + 40y2
= 15x2 + 49xy + 40y2
(viii)
(7x + 15y) (5x − 4y)
= 7x (5x − 4y) + 15y (5x − 4y)
= 35x2 − 28xy + 75xy − 60y2
= 35x2 + 47xy − 60y2
(ix)
(2a − 3b) (3a + 4b)
= 2a (3a + 4b) − 3b (3a + 4b)
= 6a2 + 8ab − 9ab − 12b2
= 6a2 − ab − 12b2
(x)
(9a − 7b) (3a − b)
= 9a (3a − b) − 7b (3a − b)
= 27a2 − 9ab − 21ab + 7b2
= 27a2 − 30ab + 7b2
Exercise – 12 B Algebraic Identities ICSE Class-8th Mathematics
Question 1 :-
Expand :
(i) (2a + b)2
(ii) (a – 2b)2
Answer :-
(i)
(2a + b)2 = (2a)2 + (b)2 + 2 × 2a × b ……[(a+b)2 = a2 + b2 + 2ab]
= 4a2 + b2 + 4ab
(ii)
(a – 2b)2 = (a)2 + (2b)2 − 2 × a × 2b ……..[(a−b)2 = a2 + b2 − 2ab]
= a2 + 4b2 − 4ab
(iii) 
(iv) 
(v)
(a+b−c)2 = (a)2 + (b)2 +(−c)2 + 2 × a × b + 2 × b × (−c) + 2 × (−c) × (a)
= a2 + b2 + c2 + 2ab − 2bc − 2ca
(vi)
(a+b+c)2 = a2 + b2 + c2 + 2ab − 2bc − 2ca
(a−b+c)2 = (a)2 + (−b)2 + (c)2 + 2 × a × −b + 2(−b)(c) + 2× c × a
= a2 + b2 + c2 − 2ab − 2bc + 2ca
(vii) 
(viii) 
Question 2 :-
Find the square of :
Answer :-
(i)
(x + 3y)2 = (x)2 + (3y)2 + 2× x × 3y
= x2 + 9y2 + 6xy
(ii)
(2x − 5y)2 = (2x)2 + (5y)2 − 2 × 2x × 5y
= 4x2 + 25y2 − 20xy
(iii) 
(iv) 
(v)
(x − 2y +1)2 = (x)2 + (−2y)2 + (1)2 + 2 × x × −2y + 2 × (−2y) ×1 + 2 × 1 × x
= x2 + 4y2 + 1 − 4xy − 4y + 2x
(vi)
(3a − 2b − 5c)2 = (3a)2 + (−2b)2 + (−5c)2 + 2 × 3a × −2b + 2 × (−2b)(−5c) + 2 × −5c × 3a
= 9a2 + 4b2 + 25c2 − 12ab + 20bc − 30ca
(vii) 
(viii) 
(ix)
(2x − 3y + z)2 = (2x)2 + (−3y)2 + (z)2 + 2 × 2x × −3y + 2(−3y) × z + 2 × z × 2x
= 4x2 + 9y2 + z2 − 12xy − 6yz + 4zx
(x) 
Question 3 :-
Evaluate:
Using expansion of (a + b)2 or (a – b)2
(i) (208)2
(ii) (92)2
(iii)(415)2
(iv) (188)2
(v) (9.4)2
(vi) (20.7)2
Answer :-
(i)
(208)2 = (200 + 8)2
= (200)2 + (8)2 + 2(200) (8)
= 40000 + 64 + 3200
= 43264
(ii)
(92)2 = (100 − 8)2
= (100)2 + (8)2 − 2(100) (8)
= 10000 + 64 − 1600
= 10064 − 1600
= 8464
(iii)
(415)2 = (400 + 15)2
= (400)2 + (15)2 + 2(400)(15)
= 160000 + 225 + 12000
= 172225
(iv)
(188)2 = (200 − 12)2
= (200)2 + (12)2 − 2(200) (12)
= 40000 + 144 − 4800
= 40144 − 4800
= 35344
(v)
(9.4)2 = (10 − 0.6)2
= (10)2 + (0.6)2 − 2 (10) (0.6)
= 100 + 0.36 − 12
= 88 + 0.36
= 88.36
(vi)
(20.7)2 = (20 + 0.7)2
= (20)2 + (0.7)2 + 2 (20) (0.7)
= 400 + 0.49 + 28
= 428 + 0.49
= 428.49
Question 4 :-
Expand :
Answer :-
(i)
(2a + b)3 = (2a)3 + (b)3 + 3×2a×b(2a + b) ……[(a + b)3 = a3 + b3 + 3ab(a + b)]
= 8a3 + b3 + 6ab (2a + b)
= 8a3 + b3 + 12a2b + 6ab2
(ii)
(a − 2b)3 = (a)3 − (2b)3 − 3×a×2b (a − 2b) ………[(a − b)3 = a3 − b3 − 3ab(a − b)]
= a3 − 8b3 − 6ab (a − 2b)
= a3 − 8b3 − 6a2b + 12ab2
(iii)
(3x − 2y)2 = (3x)3 − (2y)3 − 3 × 3x × 2y (3x − 2y)
= 27x3 − 8y3 − 18xy (3x − 2y)
= 27x3 − 8y3 − 54x2y + 36xy2
(iv)
(3x − 2y)2 = (3x)3 − (2y)3 − 3 × 3x × 2y (3x − 2y)
= 27x3 − 8y3 − 18xy (3x − 2y)
= 27x3 − 8y3 − 54x2y + 36xy2
(v)
(vi)
Question 5 :-
Find the cube of :
Answer :-
(i)
(a + 2)3 = (a)3 + (2)3 + 3×a×2(a + 2)
= a3 + 8 + 6a(a+ 2)
= a3 + 8 + 6a2 + 12a
= a3 + 6a2 + 12a+ 8
(ii)
(2a − 1)3 = (2a)3 − (1)3 − 3×2a×1(2a − 1)
= 8a3 − 1 − 6a (2a − 1)
= 8a3 − 1 − 12a2 + 6a
= 8a3 − 12a2 + 6a − 1
(iii)
(2a + 3b)3 = (2a)3 + (3b)3 + 3×2a×3b(2a + 3b)
= 8a3 + 27b3 + 18ab (2a + 3b)
= 8a3 + 27b3 + 36a2b + 54ab2
= 8a3 + 36a2b + 54ab2 + 27b3
(iv)
(3b − 2a)3 = (3b)3 − (2a)3 − 3×3b×2a(3b − 2a)
= 27b3 − 8a3 − 18ab (3b − 2a)
= 27b3 − 8a3 − 54ab2 + 36a2b
= 27b3 − 54b2a + 36ba2 − 8a3
(v)
(vi)
Solved Questions of Exe-12 C for ICSE Class-8th Mathematics
Question 1 :-
If a +b = 5 and ab = 6; find a2 + b2
Answer :-
(a + b)2 = a2 + b2 + 2ab
⇒ (5)2 = a2 + b2 + 2×6
⇒ 25 = a2 + b2 + 12
⇒ 25 − 12 = a2 + b2
⇒ 13 = a2 + b2
∴ a2 + b2 = 13
Question 2 :-
If a – b = 6 and ab = 16; find a2 + b2
Answer :-
(a − b)2 = a2 + b2 − 2ab
⇒ (6)2 = a2 + b2 − 2×16
⇒ 36 = a2 + b2 − 32
⇒ 36 + 32 = a2 + b2
⇒ 68 = a2 + b2
∴ a2 + b2 = 68
Question 3 :-
If a2 + b2 = 29 and ab = 10 ; find :
(i) a + b
(ii) a – b
Answer :-
⇒ (a + b)2 = 29 + 2×10
⇒ (a + b)2 = 29 + 20
⇒ (a + b)2 = 49
⇒ a + b = √49
⇒ a + b = ±7
⇒ (a − b)2 = 29 − 2×10
⇒ (a − b)2 = 29 − 20
⇒ (a − b)2 = 9
⇒ a − b = √9
⇒ a − b = ±3
Question 4 :-
If a2 + b2 = 10 and ab = 3; find :
(i) a – b
(ii) a + b
Answer :-
(i)
(a − b)2 = a2 + b2 − 2ab
⇒ (a − b)2 = 10 − 2×3
⇒ (a − b)2 = 10 − 6
⇒ (a − b)2 = 4
⇒ (a − b) = √4
⇒ a − b = ±2
(ii)
(a + b)2 = a2 + b2 + 2ab
⇒ (a + b)2 = 10 + 2×3
⇒ (a + b)2 = 10 + 6
⇒ (a + b)2 = 16
⇒ (a + b) = √16
⇒ a + b = ±4
Question 5 :-
Answer :-
Question 6 :-
Answer :-
Question 7 :-
Answer :-
Question 8 :-
Answer :-
Question 9 :-
If a + b + c = 10 and a2 + b2 + c2 = 38; find ab + bc + ca
Answer:-
a + b + c = 10
⇒ (a + b + c)2 = (10)2
⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 100
⇒ 38 + 2(ab + bc + ca) = 100
⇒ 2(ab + bc + ca) = 100 − 38
⇒ 2(ab + bc + ca) = 62
⇒ (ab + bc+ ca) =62/2
⇒ ab + bc + ca = 31
Question 10 :-
Find a2 + b2 + c2 ; if a + b + c = 9 and ab + bc + ca = 24
Answer :-
a+ b + c = 9
⇒ (a + b + c)2 = (9)2
⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 81
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81
⇒ a2 + b2 + c2 + 2×24 = 81
⇒ a2 + b2 + c2 + 48 = 81
⇒ a2 + b2 + c2 = 81 − 48
⇒ a2 + b2 + c2 = 33
Question 11 :-
Find a + b + c; if a2 + b2 + c2 = 83 and ab + bc + ca = 71
Answer :-
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a + b + c)2 = 82 + 2(ab + bc + ca)
⇒ (a + b+ c)2 = 83 + 2×71
⇒ (a + b + c)2 = 83 + 142
⇒ (a + b + c)2 = 225
⇒ a + b + c = √225
⇒ a + b +c = ±15
Question 12 :-
If a + b = 6 and ab=8; find a3 + b3
Answer :-
a + b = 6
⇒ (a + b)3 = (6)3
⇒ a3 + b3 + 3ab (a + b) = 216
⇒ a3 + b3 + 3×8 (6) = 216
⇒ a3 + b3 + 144 = 216
⇒ a3 + b3 = 216 − 144
⇒ a3 + b3 = 72
Question 13 :-
If a – b = 3 and ab = 10; find a3 – b3
Answer :-
a − b = 3
⇒ (a − b)3 = (3)3
⇒ a3 − b3 − 3ab (a − b) = 27
⇒ a3 − b3 − 3×10 (3) = 27
⇒ a3 − b3 − 90 = 27
⇒ a3 − b3 = 27 + 90
⇒ a3 − b3 = 117
Question 14 :-
Answer :-
Question 15 :-
Answer :-
Question 16 :-
Answer :-
(i)
(ii)
Question 17 :-
Answer :-
(i)
(ii)
Question 18 :-
The sum of the squares of two numbers is 13 and their product is 6. Find:
(i) the sum of the two numbers.
(ii) the difference between them.
Answer :-
Let x and y be the two numbers, then,
x2 + y2 = 13 and xy = 6
(i) (x + y)2 = x2 + y2 + 2xy
= 13 + 2×6
= 13 + 12
= 25
∴ x + y = ± √25 = ±5
(ii) (x − y)2 = x2 + y2 − 2xy
= 13 − 12
= 1
∴ x − y = ±1
Algebraic Identities ICSE Class-8th Maths Exe- 12 D
Question 1 :-
Evaluate:
Answer :-
(i) 
(ii)
(2a + 0.5) (7a − 0.3)
= 2a (7a − 0.3) + 0.5 (7a − 0.3)
= 14a2 − 0.6a + 3.5a − 0.15
= 14a2 + 2.9a − 0.15
(iii)
(9 − y) (7 + y) = 9(7 + y) − y (7 + y)
= 63 + 9y − 7y − y2
= 63 + 2y − y2
(iv)
(2 − z) (15 − z) = 2(15 − z) −z(15 − z)
= 30 − 2z − 15z + z2
= 30 − 17z + z2
(v)
(a2 + 5) (a2 − 3) = a2 (a2 − 3) + 5 (a2 − 3)
= a4 − 3a2 + 5a2 − 15
= a4 + 2a2 − 15
(vi)
(4 − ab) (8 + ab) = 4 (8 + ab) − ab (8 + ab)
= 32 + 4ab − 8ab − a2b2
= 32 − 4ab − a2b2
(vii)
(5xy − 7) (7xy + 9) = 5xy (7xy + 9) − 7 (7xy + 9)
= 35x2y2 + 45xy − 49xy − 63
= 35x2y2 − 4xy − 63
(viii)
(3a2 − 4b2) (8a2 − 3b2)
= 3a2 (8a2 − 3b2) − 4b2 (8a2 − 3b2)
= 24a4 − 9a2b2 − 32a2b2 + 12b4
= 24a4 − 41a2b2 + 12b4
Question 2 :-
Evaluate:
Answer :-
(i)
(ii)
(iii)
(6 − 5xy) (6 + 5xy)
= (6)2 − (5xy)2
= 36 − 25x2y2 ………[ ∵ (a − b) (a + b) = a2b2]
(iv)
(v)
(4x2 − 5y2) (4x2 + 5y2)
= (4x)2 − (5y2)2
= 16x4 − 25y4 ……[ ∵ (a − b) (a + b) = a2 − b2]
(vi)
(1.6x + 0.7y) (1.6x − 0.7y)
= (1.6x)2 − (0.7y)2 ……[ ∵ (a − b) (a +b) = a2 − b2]
= 2.56x2 − 0.49y2
(vii)
(m + 3) (m − 3) (m2 + 9)
= (m)2 − (3)2 (m2 + 9) ……[ ∵ (a − b) (a + b) = a2 − b2]
= (m2 − 9) (m2 + 9)
= (m2)2 − 92
= m4 − 81
(viii)
(3x + 4y) (3x − 4y) (9x2 + 16y2)
= [(3x)2 − (4y)2] (9x2 + 16y2) …..[ ∵ (a − b) (a + b) = a2 − b2]
= (9x2 − 16y2) (9x2 + 16y2)
= (9x2)2 − (16y2)2 ……[ ∵ (a − b) (a + b) = a2 − b2]
= 81x4 − 256y4
(ix)
(a + bc) (a − bc) (a2 + b2c2)
= [a2 − (bc)2] (a2 + b2c2) …….[ ∵ (a − b) (a + b) = a2 − b2]
= (a2 − b2c2) (a2 + b2c2)
= (a2)2 − (b2c2)2 …….[ ∵ (a − b) (a + b) = a2 − b2]
= a4 − b4c4
(x)
203 × 197
= (200 + 3) (200 − 3)
= (200)2 − (3)2 ……[ ∵ (a − b) (a + b) = a2 − b2]
= 40000 − 9
= 39991
(xi)
20.8 × 19.2
= (20 + 0.8) (20 − 0.8)
= (20)2 − (0.8)2 ……[ ∵ (a − b) (a + b) = a2 − b2]
= 400 − 0.64
= 399.36
Question 3 :-
Find the square of :
Answer :-
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(607)2 = (600 + 7)2
= (600)2 + (7)2 + 2 (600) (7)
= 360000 + 49 + 8900
= 368449
(vii)
(319)2 = (400 − 9)2
= (400)2 + 92 + 2 (400) (9)
= 160000 + 81 − 7200
= 152881
(vii)
(9.7)2 = (10 − 0.3)2
= (10)2 + (0.3)2 − 2 (10) (0.3)
= 100 + 0.9 − 6
= 100.09 − 6.00
= 94.09
Question 4 :-
Answer :-
(i)
= 4 − 2
= 2
(ii)
= 4 − 2
= 2
Question 5 :-
Answer :-
(i)
= (5)2 + 2
= 25 + 2
= 27
(ii)
= (27)2 − 2
= 729 − 2
= 727
(iii)
Question 6 :-
If a2 + b2 = 41 and ab = 4, find :
(i) a – b
(ii) a + b
Answer :-
(i)
(a − b)2 = a2 + b2 − 2ab
= 41 − 2(4)
= 41 − 8
= 33
∴ a − b = √33
(ii)
(a + b)2 = a2 + b2 + 2ab
= 41 + 2(4)
= 41 + 8
= 49
⇒ (a + b)2 = 49
∴ a + b = 7
Question 7 :-
Answer :-
(i)
= (8)2 − 2
= 64 − 2
= 62
(ii)
= (62)2 − 2
= 3844 − 2
= 3842
Question 8 :-
Answer :-
(i)
= (5)2 + 2
= 25 + 2
= 27
(ii)
= (27)2 − 2
= 729 − 2
= 727
Question 9 :-
Expand :
(i) (3x – 4y + 5z)2
(ii) (2a – 5b – 4c)2
(iii) (5x + 3y)3
(iv) (6a – 7b)3
Answer :-
(i)
(3x – 4y + 5z)2
= (3x)2 + (– 4y)2 + (5z)2 + 2(3x) (– 4y) + 2(– 4y) (5z) + 2(5z) (3x)
= 9x2 + 16y2 + 25z2 – 24xy – 40yz + 30zx
(ii)
(2a – 5b – 4c)2
= (2a)2 + (–5b)2 + (–4c)2 + 2 (2a) (–5b) + 2 (–5b) (–4c) + 2 (–4c) (2a)
= 4a2 + 25b2 + 16c2 – 20ab + 40bc – 16ca
(iii)
(5x + 3y)3
= (5x)3 + (3y)3 + 3 (5x) (3y) (5x + 3y)
= 125x3 + 27y3 + 45xy (5x + 3y)
= 125x3 + 27y3 + 225x2y + 135xy2
(iv)
(6a – 7b)3
= (6a)3 – (7b)3 – 3(6a) (7b) (6a – 7b)
= 216a3 – 343b3 – 126ab (6a – 7b)
= 216a3 – 343b3 – 756a2b + 882ab2
= 216a3 – 756a2b + 882ab2 – 343b3
Question 10 :-
If a + b + c = 9 and ab + bc + ca = 15, find: a2 + b2 + c2.
Answer :-
Since (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
Question 11 :-
If a + b + c = 11 and a2 + b2 + c2 = 81, find ab + bc + ca.
Answer :-
Since (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
∴ (11)2 = 81 + 2 (ab + bc + ca)
∴ 2(ab + bc + ca) = 121 − 81 = 40
ab + bc + ca = 40/2
⇒ ab + bc + ca = 20
Question 12 :-
If 3x – 4y = 5 and xy = 3, find : 27x3 – 64y3.
Answer :-
27x3 – 64x3 = (3x)3 – (4y)3
= (3x − 4y)3 (3x − 4y)3 + 3 (3x) (4y) (3x − 4y) ……[ ∵ a3 − b3 = (a − b)3 + 3ab(a − b)]
= (5)3 + 36(xy) (3x − 4y)
= 125 + 36 (3) (5)
= 125 + 540
= 665
Question 13 :-
If a + b = 8 and ab = 15, find : a3 + b3.
Answer :-
a3 + b3 = (a + b)3 − 3ab (a + b)
= (8)3 − 3(15) (8)
= 512 − 360
= 152
Question 14 :-
If 3x + 2y = 9 and xy = 3, find : 27x3 + 8y3
Answer :-
27x3 + 8y3 = (3x)3 + (2y)3
= (3x + 2y)3 − 3.3x . 2y (3x + 2y)
= (3x − 2y)3 − 18xy (3x + 2y)
= (9)3 − 18(3) (9)
= 729 − 486
= 243
Question 15 :-
If 5x – 4y = 7 and xy = 8, find : 125x3 – 64y3
Answer :-
125x3 − 64y3 = (5x)3 − (4y)3
= (5x − 4y)3 + 3(5x) (4y) (5x − 4y)
= (5x − 4y)3 + 60xy (5x − 4y)
= (7)3 + 60 (8) (7)
= 343 + 3360
= 3703
Question 16 :-
The difference between two numbers is 5 and their products is 14. Find the difference between their cubes.
Answer :-
Let x and y be two numbers, then x – y = 5 and xy = 14
∴ x3 − y3 = (x − y)3 + 3xy(x − y)
= (5)3 + 3 × 14 × 5
= 125 + 210
= 335
— End of Algebraic Identities Solutions :–
Return to – Concise Selina Maths Solutions for ICSE Class -8
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