c) (x – 1)^3 – 27
c) (x – 1)3 – 27
c) (x – 1)3 – 27
= (x – 1)3 – 33
= (x ‒ 1 ‒ 3)[(x ‒ 1)2 + (x ‒ 1).3 + 32]
= (x ‒ 4)(x2 ‒ 2x + 1 + 3x ‒ 3 + 9)
= (x ‒ 4)[x2 + (‒2x + 3x) + 1 ‒ 3 + 9]
= (x ‒ 4)(x2 + x + 7).