a^4+b^4+c^4-2(ab)^2-2(ac)^2-2(bc)^2分解因式,
1个回答

a^4+b^4+c^4-2(ab)^2-2(ac)^2-2(bc)^2

=(a^4-2a²b²+b^4)+c^4-2a²c²-2b²c²

=(a²-b²)²+c^4-c²(2a²-2b²)

=(a+b)²(a-b)²+(c²)²-c²(a²+2ab+b²+a²-2ab+b²)

=(a+b)²(a-b)²+(c²)²-c²[(a+b)²+(a-b)²]

设(a+b)²=x,(a-b)²=y,c²=z

原式

=xy+z²-z(x+y)

=xy-zx+z²-zy

=x(y-z)-z(y-z)

=(x-z)(y-z)

=[(a+b)²-c²][(a-b)²-c²]

=(a+b-c)(a+b+c)(a-b+c)(a-b-c)