Hoặc
b) (x + 1)3 + (x – 1)3;
b) (x + 1)3 + (x – 1)3
= x3 + 3x2 + 3x + 1 + x3 – 3x2 + 3x – 1
= (x3 + x3) + (3x2 – 3x2) + (3x + 3x) + (1 – 1)
= 2x3 + 6x = 2x(x2 + 3);