(x – 1)( x 2 + 5x – 2) – ( x 3 – 1) = 0
⇔ (x – 1)( x 2 + 5x – 2) – (x – 1)( x 2 + x + 1) = 0
⇔ (x – 1)[( x 2 + 5x – 2) – ( x 2 + x + 1)] = 0
⇔ (x – 1)( x 2 + 5x – 2 – x 2 – x – 1) = 0
⇔ (x – 1)(4x – 3) = 0 ⇔ x – 1 = 0 hoặc 4x – 3 = 0
x – 1 = 0 ⇔ x = 1
4x – 3 = 0 ⇔ x = 0,75
Vậy phương trình có nghiệm x = 1 hoặc x = 0,75