lf X+y+z=0 show that .
x^3+y^3+z^3=3xyz
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We know that,
x³+y³+z³-3xyx=(x+y+z)(x²+y²+z²-xy-yz-zx)
It is given that,
x+y+z=0
so,
x³+y³+z²-3xyz=(0)(x²+y²+z²-xy-yz-zx)
x³+y³+z³-3xyx=0 ( anything multiplies with zero is zero ).
Hence,
x³+y³+z³-3xyz=0
x³+y³+z³=3xyz
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