可以幫我因式分解這兩題嗎? (1)(x^2-y^2)^2-8(x^2+y^2)+16 (2)(2x+1)(x^2-x+3)+2x(x-6) ps:我要過程?

(1)(x^2－y^2)^2－8(x^2+y^2)+16
=(x^2－y^2)^2－8(x^2－y^2+2y^2)+16
=(x^2－y^2)^2－8(x^2－y^2)－16y^2+16
=(x^2－y^2)^2－8(x^2－y^2)+16－16y^2
=(x^2－y^2－4)^2－(4y)^2
=(x^2－y^2－4y－4)(x^2－y^2+4y－4)
=[x^2－(y+2)^2]*[x^2－(y－2)^2]
=(x－y－2)(x+y+2)(x－y+2)(x+y－2)
(2)(2x+1)(x^2－x+3)+2x(x－6)
=(2x^3－2x^2+6x+x^2－x+3)+(2x^2－12x)
=2x^3+x^2－7x+3
=(2x^3－x^2)+(2x^2－x)+(－6x+3)
=x^2(2x－1)+x(2x－1)－3(2x－1)
=(x^2+x－3)(2x－1)

(1) (x²-y²)²-8(x²+y²)+16
=(x²-y²-4)(x²-y²-4)
(2)(2x+1)(x²-x+3)+2x(x-6)
=2x³-2x²+6x+x²-x+3+2x²-12x
=2x³+x²-7x+3
=2x³+x²-x-6x+3
=x(2x²+x-1)-3(2x-1)
=x(2x-1)(x+1)-3(2x-1)
=(2x-1)(x²+x-3)
算好久0.0請版主自己再確認是否有錯誤喔!!!希望能幫到你^^