Question

若x+y+z=0 试求(x²-y²-z²)²-4y²z² 请尽量用通俗的办法哈

Answer 1

因为x+y+z=0
所以x+y=-z x+z=-y

(x²-y²-z²)²-4y²z²
=[(x+y)(x-y)-z²]²-4y²z²
=[-z(x-y)-z²]²-4y²z²
=[-z(x-y+z)]²-4y²z²
=[-z(-y-y)]²-4y²z²
=(2zy)²-4y²z²
=4y²z²-4y²z²
=0

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Answer 2

平方差
原式=(x²-y²-z²+2yz)(x²-y²-z²-2yz)
=[x²-(y²-2yz+z²)][x²-(y²+2yz+z²)]
=[x²-(y-z)²][x²-(y+z)²]
=(x+y-z)(x-y+z)(x+y+z)(x-y-z)
x+y+z=0
所以原式=0

Answer 3

因为x+y+z=0
所以y+z=-x

(x²-y²-z²)²-4y²z² =x^4-2*x^2*(y²+z²)+(y²+z²)^2-4y²z²
=x^4-2*x^2*(y²+z²)+(y²-z²)^2
=x^4-2*x^2*(y²+z²)+(y-z)^2(y+z)^2

=x^4-2*x^2*(y²+z²)+(y-z)^2x^2
=x^4-x^2(2y²+2z²-y²-z²=2yz)
=x^4-x^2(y+z)^2
=x^4-x^4
=0