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).