If X+ Y + Z = 0 show that x cube + y cube + z cube = 3 xyz.
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We know that,x^3+y^3+z^3-3xyz
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
Given,x+y+z=0. Then,x^3+y^3+z^3-3xyz
=0×(x^2+y^2+z^2-xy-yz-zx)
=0
So,x^3+y^3+z^3=3xyz. (PROVED)
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
Given,x+y+z=0. Then,x^3+y^3+z^3-3xyz
=0×(x^2+y^2+z^2-xy-yz-zx)
=0
So,x^3+y^3+z^3=3xyz. (PROVED)
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