Question

把2(x^2+y^2)^2-8xy(x^2+y^2)+8x^2y^2因式分解

Answer 1

2(x^2+y^2)^2-8xy(x^2+y^2)+8x^2y^2

=2[(x^2+y^2)^2-4xy(x^2+y^2)+4x^2y^2]
=2(x²+y²-2xy)²
=2(x-y)^4

追问

• 最后是2（x-y)^2吧

追答

是2（x-y)^4

=2(x²+y²-2xy)²
=2[(x-y)²]²
=2(x-y)^4

2(x²+y²)²-8xy(x²+y²)+8x²y²
=2[(x²+y²)²-4xy(x²+y²)+4x²y²]
=2[(x²+y²)²-2*2xy(x²+y²)+(2xy)²]
=2(x²+y²-2xy)²......把上一步(x²+y²)看作a,(2xy)看作b,这里就是一个完全平方,(a²-2ab+b²)
=2[(x-y)²]²............上一步的x²+y²-2xy又是一个完全平方,
=2(x-y)^4.............最后得到的是4次方

这样能看懂吗

Answer 2

原式=2[(x^2+y^2)^2-4xy(x^2+y^2)+4x^2y^2]
=2[(x^2+y^2)-2xy]^2
=2[(x-y)^2]^2
=2(x-y)^4 。