Question

因式分解x4+y4+z4-2x2y2-2x2z2-2y2z2

x,y,后的2为平方。求详细步骤

Answer 1

x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2
=x^4-2x^2y^2+y^4+z^4-2x^2z^2-2y^2z^2
=(x^4-2x^2y^2+y^4)+(z^4-2x^2z^2-2y^2z^2)
=(x^2-y^2)^2+z^4-2y^2z^2+2z^2x^2-4z^2x^2
=(x^2-y^2)^2+2z^2(x^2-y^2)+z^4-4z^2x^2
=(x^2-y^2+z^2)^2-4z^2x^2
=(x^2-y^2+z^2+2zx)(x^2-y^2+z^2-2zx)
=[(x+z)^2-y^2][(x-z)^2-y^2]
=(x+y+z)(x-y+z)(x+y-z)(x-y-z)

Answer 2

x^4 + y^4 + z^4 - 2x^2y^2 - 2x^2z^2 - 2y^2z^2
=[z^2 - (x+y)^2][z^2 - (x-y)^2]
=(z + x + y)(z - x - y)(z + x - y)(z - x + y)