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)