Xét riêng (x + y)^4 = [(x + y)^2]^2 = [x^2+2xy+y^2]^2 = x^4 +4x^2y^2 + y^4 + 4x^3y + 2x^2y^2+4xy^3
Vậy (x + y)^4 +x^4 + y^4 = x^4 +4x^2y^2 + y^4 + 4x^3y + 2x^2y^2+4xy^3+ x^4 + y^4
= 2x^4 + 2y^4 + 6x^2y^2 + 4x^3y + 4xy^3
= 2(x^4 + y^4 + 3x^2y^2 +2 x^3y + 2xy^3)
= 2(x^4 + y^4 + x^2y^2 + 2x^3y + 2xy^3 + 2x^2y^2)
= 2(x^2 + xy + y^2)^2