Xét riêng (x + y)4 = [(x + y)2]2
= [x2+2xy+y2]2
= x4 +4x2y2 + y4 + 4x3y + 2x2y2+4xy3
Vậy (x + y)4 +x4 + y4 = x4 +4x2y2 + y4 + 4x3y + 2x2y2+4xy3+ x4 + y4
= 2x4 + 2y4 + 6x2y2 + 4x3y + 4xy3
= 2(x4 + y4 + 3x2y2 +2x3y + 2xy3)
= 2(x4 + y4 + x2y2 + 2x3y + 2xy3 + 2x2y2)
= 2(x2 + xy + y2)2