因式分解35分:x^4+y^4+2(x+y)^4-(x^2+y^2)^2-4xy(x^2+xy+y^2)

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因式分解35分:x^4+y^4+2(x+y)^4-(x^2+y^2)^2-4xy(x^2+xy+y^2)

原式=2(x+y)^4-4x^3y-4xy^3-6x^2y^2
=2(x+y)^4-4xy(x^2+2xy+y^2)+2x^2y^2
=2[(x+y)^2-xy]^2
=2(x^2+xy+y^2)^2
=2x^4+2y^4+4x^3y+4xy^3+6x^2y^2
即需要注意将式子当做一个整体的思想.

x^4+y^4+2(x+y)^4-(x²+y²)²-4xy(x²+xy+y²)
=2(x+y)^4-(x^4+2x²y²+y^4)+x^4+y^4-(4x³y+4x²y²+4xy³)
=2(x+y)^4+(x^4-x^4)+(y^4-y^4)-4x³y-6x&su...

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x^4+y^4+2(x+y)^4-(x²+y²)²-4xy(x²+xy+y²)
=2(x+y)^4-(x^4+2x²y²+y^4)+x^4+y^4-(4x³y+4x²y²+4xy³)
=2(x+y)^4+(x^4-x^4)+(y^4-y^4)-4x³y-6x²y²-4xy³+(2x²y²-2x²y²)
=2(x+y)^4-4x³y-8x²y²-4xy³+2x²y²
=2(x+y)^4-4xy(x²+2xy+y²)+2(xy)²
=2(x+y)^4-4xy(x+y)²+2(xy)²
=2[(x+y)^4-2xy(x+y)²+(xy)²],令a=x+y,b=xy
=2[(a²)²-2a²b+b²)
=2(a²-b)²
=2[(x+y)²-xy]²

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x^4+y^4+2(x+y)^4-(x²+y²)²-4xy(x²+xy+y²)
=2(x+y)^4-(x^4+2x²y²+y^4)+x^4+y^4-(4x³y+4x²y²+4xy³)
=2(x+y)^4+(x^4-x^4)+(y^4-y^4)-4x³y-6x&su...

全部展开

x^4+y^4+2(x+y)^4-(x²+y²)²-4xy(x²+xy+y²)
=2(x+y)^4-(x^4+2x²y²+y^4)+x^4+y^4-(4x³y+4x²y²+4xy³)
=2(x+y)^4+(x^4-x^4)+(y^4-y^4)-4x³y-6x²y²-4xy³+(2x²y²-2x²y²)
=2(x+y)^4-4x³y-8x²y²-4xy³+2x²y²
=2(x+y)^4-4xy(x²+2xy+y²)+2(xy)²
=2(x+y)^4-4xy(x+y)²+2(xy)²
=2[(x+y)^4-2xy(x+y)²+(xy)²],令a=x+y,b=xy
=2[(a²)²-2a²b+b²)
=2(a²-b)²
=2[(x+y)²-xy]²
=2(x²+xy+y²)² ~~~楼上回答不错~~~

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