Question

分解因式1)(x+y)^4+(x^2-y^2)^2+(x-y)^4(2)x²-y²+3x-y+2(3

Answer 1

1)(x+y)^4+(x^2-y^2)^2+(x-y)^4

=(x+y)^4+(x+y)^2(x-y)^2+(x-y)^4

=(x+y)^4+2(x+y)^2(x-y)^2+(x-y)^4-(x+y)^2(x-y)^2

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

=(2x^2+2y^2)^2-(x^2-y^2)^2

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

=(3x^2+y^2)(x^2+3y^2)

(2)

x^2-y^2+3x-y+2

=(x^2+3x+9/4)-(y^2+y+1/4)

=(x+3/2)^2-(y+1/2)^2

=(x+3/2-y-1/2)(x+3/2+y+1/2)

=(x-y+1)(x+y+2).

(3)

x^2+xy-2y^2-x+7y-6

=（x+2y)(x-y)-x+7y+2*(-3)

=(x+2y-3)(x-y+2)