(i) (x + 4)(x + 10): Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, we have: (x + 4)(x + 10) = x2 + (4 + 10)x + (4 x 10) = x2 + 14x + 40 (ii) (x + 8)(x – 10): Here, a = 8 and b = (–10) ∴ Using (x + a)(x + b) = x2 + (a + b)x + ab, we have: (x + 8)(x – 10) = x2 + [8 + (–10)]x + [8 x (–10)] = x2 + [–2]x + [–80] = x2 – 2x – 80 2. Evaluate the following products without multiplying directly: (i) 103 x 107 (ii) 95 x 96 (iii) 104 x 96 (i) We have 103 x 107 = (100 + 3)(100 + 7) = (100)2 + (3 + 7) x 100 + (3 x 7) [Using (x + a)(x + b) = x2 + (a + b)x + ab] = 10000 + (10) x 100 + 21 = 10000 + 1000 + 21 = 11021 (ii) We have 95 x 96 = (100 – 5)(100 – 4) = (100)2 + [(–5) + (–4)] x 100 + [(–5) x (–4)] [Using (x + a)(x + b) = x2 + (a + b)x + ab] = 10000 + [–9] x 100 + 20 = 10000 + (–900) + 20 = 9120 (iii) We have 104 x 96 = (100 + 4)(100 – 4) = (100)2 – (4)2 [Using (a + b)(a – b) = a2 – b2] = 10000 – 16 = 9984 3. Factorise the following using appropriate identities. 4. Expand each of the following, using suitable identities: (i) (x + 2y + 4z)2 We have (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx ∴ (x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x) = x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx (ii) (2x – y + z)2 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (2x – y + z)2 = (2x)2 + (–y)2 + (z)2 + 2(2x)(–y) + 2(–y)(z) + 2(z)(2x) = 4x2 + y2 + z2 – 4xy – 2yz + 4zx (iii) (–2x + 3y + 2z)2 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (–2x + 3y + 2z)2 = (–2x)2 + (3y)2 + (2z)2 + 2(–2x)(3y) + 2(3y)(2z) + 2(2z)(–2x) = 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx (iv) (3a – 7b – c)2 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (3a – 7b – c)2 = (3a)2 + (–7b)2 + (–c)2 + 2(3a)(–7b) + 2(–7b)(–c) + 2(–c)(3a) = 9a2 + 49b2 + c2 + (–42ab) + (14bc) + (–6ca) = 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca (v) (–2x + 5y – 3z)2 Using (x + y + z)2 =x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (–2x + 5y – 3z)2 = (–2x)2 + (5y)2 + (–3z)2 + 2(–2x)(5y) + 2(5y)(–3z) + 2(–3z) (–2x) =4x2 + 25y2 + 9z2 + [–20xy] + [–30yz] + [12zx] =4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx 5. Factorise: (i)4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz = (2x)2 + (3y)2 + (–4z)2 + 2(2x)(3y) + 2(3y)(–4z) + 2(–4z)(2x) = (2x + 3y – 4z)2 [Using Identity V] = (2x + 3y – 4z)(2x + 3y – 4z) 6. Write the following cubes in expanded form: (i) (2x + 1)3 (ii) (2a – 3b)3 Using Identity VI and Identity VII, we have (x + y)3 = x3 + y3 + 3xy (x + y), and (x – y)3 = x3 – y3 – 3xy (x – y). (i) (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1)[(2x) + (1)] = 8x3 + 1 + 6x[2x + 1] [Using Identity VI] = 8x3 + 1 + 12x2 + 6x = 8x3 + 12x2 + 6x + 1 (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)[(2a) – (3b)] = 8a3 – 27b3 – 18ab (2a – 3b) [Using Identity VII] = 8a3 – 27b3 – [36a2b – 54ab2] = 8a3 – 27b3 – 36a2b + 54ab2 7. Evaluate the following using suitable identities: (i) (99)3 (ii) (102)3 (iii) (998)3 (i) (99)3 We have 99 = 100 – 1 ∴ 993 = (100 – 1)3 = 1003 – 13 – 3(100)(1)(100 – 1) = 1000000 – 1 – 300(100 – 1) = 1000000 – 1 – 30000 + 300 = 1000300 – 30001 = 970299 (ii) (102)3 We have 102 = 100 + 2 ∴ (102)3 = (100 + 2)3 = (100)3 + (2)3 + 3(100)(2) 100 + 2 = 1000000 + 8 + 600[100 + 2] = 1000000 + 8 + 60000 + 1200 = 1061208 (iii) (998)3 We have 998 = 1000 – 2 ∴ (999)3 = (1000 – 2)3 = (1000)3 – (2)3 – 3(1000)(2)[1000 – 2] = 1000000000 – 8 – 6000[1000 – 2] = 1000000000 – 8 – 6000000 – 12000 = 994011992 8. Factorise each of the following: (i) 8a3 + b3 + 12a2b + 6ab2 (ii) 8a3 – b3 – 12a2b + 6ab2 (iii) 27 – 125a3 – 135a + 225a2 (iv) 64a3 – 27b3 – 144a2b + 108ab2 (i)8a3 + b3 + 12a2b + 6ab2 = (2a)3 + (b)3 + 6ab(2a + b) = (2a)3 + (b)3 + 3(2a)(b)(2a + b) = (2a + b)3 [Using Identity VI] = (2a + b)(2a + b)(2a + b) (ii)8a3 – b3 – 12a2b + 6ab2 = (2a)3 – (b)3 – 3(2a)(b)(2a – b) = (2a – b)3 [Using Identity VII] = (2a – b)(2a – b)(2a – b) (iii) 27 – 125a3 – 135a + 225a2 = (3)3 – (5a)3 – 3(3)(5a)[3 – 5a] = (3 – 5a)3 [Using Identity VII] = (3 – 5a)(3 – 5a)(3 – 5a) (iv) 64a3 – 27b3 – 144 a2b + 108 ab2 = (4a)3 – (3b)3 – 3(4a)(3b)[4a – 3b] = (4a – 3b)3 [Using Identity VII] = (4a – 3b)(4a – 3b)(4a – 3b) <!--[endif]--> 9. Verify: (i) x3 + y3 = (x + y)(x2 – xy + y2) (ii) x3 – y3 = (x – y)(x2 + xy + y2) (i) R.H.S. = (x + y)(x2 – xy + y2) = x(x2 – xy + y2) + y(x2 – xy + y2) = x3 – x2y + xy2 + x2y – xy2 + y3 = x3 + y3 = L.H.S. (ii) R.H.S. = (x – y)(x2 + xy + y2) = x(x2 + xy + y2) – y(x2 + xy + y2) = x3 + x2y + xy2 – x2y – xy2 – y3 = x3 – y3 = L.H.S. 10. Factorise each of the following: (i) 27y3 + 125z3 (ii) 64m3 – 343n3 REMEMBER I. x3 + y3 = (x + y)(x2 + y2 – xy) II. x3 – y3 = (x – y)(x2 + y2 + xy) (i) Using the identity (x3 + y3) = (x + y)(x2 – xy + y2), we have 27y3 + 125z3 = (3y)3 + (5z)3 = (3y + 5z) [(3y)2 – (3y)(5z) + (5z)2] = (3y + 5z)(9y2 – 15yz + 25z2) (ii) Using the identity x3 – y3 = (x – y)(x2 + xy + y2), we have 64m3 – 343n3 = (4m)3 – (7n)3 = (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2] = (4m – 7n)(16m2 + 28mn + 49n2)