Factorise:
(i) 1 + 64
(i) Given that: 1 + 64x3
= 1 + (4x)3
= (1 + 4x)[(1)2 − (1)(4x) + (4x)2]
[∵ a3 + b3 = (a + b)(a2 – ab + b2) ]
= (1 + 4x)(1 – 4x + 16x2)
(ii) Given that: a3 − 2√2b3
= (a)3 − (√2b)3
= (a − √2b) [(a)2 + (a)(√2b) + (√2b)2]
[∵ a3 – b3 = (a + b)(a2 + ab + b2) ]
= (a − √2b)(a2 + √2ab + 2b2)