(i) (a + 2b)3 – (a – 2b)3
= (a + 2b – a + 2b) [(a + 2b)2 + (a – 2b)2 + (a + 2b) (a – 2b)]
= (4b)[a2 + 4b2 + 4ab + a2 + 4b2 – 4ab + a2 – 4b2]
= (4b)[3a2 + 4b2]
(ii) a3 + b3 + c(a2 – ab + b2) = (a + b)(a + b2 – ab) + c(a2 – ab + b)
= (a2 + b2 – ab)[a + b + c]
(iii) x6 – 7x3 – 8 = x6 – (8 – 1)x3 – 8 (∵ 8 = 1 × 8)
= x6 – 8x3 + x3 – 8 = x3 (x3 – 8) +1(x3 – 8) = (x3 – 8)(x3 + 1)
=[(x)3 – (2)3][(x)3 + (1)3] = (x – 2)(x2 + 4 + 2x)(x + 1)(x2 + 1 – x)
(iv) x3 – 3x2 + 3x + 7
माना f(x) = x3 – 3x2 + 3x + 7x + 1 = 0
x = -1 रखने पर,
= (-1)3 – 3(-1)2 + 3 × (-1) +7
= -1 – 3 – 3 + 7 = -7 + 7 = 0
∵ (x + 1), f(x) का एक गुणनखण्ड है।
अतः
अत: x3 – 3x2 + 3x + 7 के गुणनखण्ड (x + 1)(x2 – 4x + 7) है।
