prove that 2x³+2y³+2z³-6xyz=(x+y+z)[(x-y)²+(y-z)²+(z-x)²]
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x^3 +y^3 +z^3- 3xyz = (x+y+z)(x^2 +y^2 +z^2-xy -yz-zx)
=(x+y+z)×1/2[(x-y)^2+(z-x)^2+(y-z)^2
as (x-y)^2 =x^2 +y^2-2xy
2(x^3+y^3+z^3 -3xyz)=2×1/2 (x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]
2x^3+2y^3+2z^3 -6xyz=(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2] proved
=(x+y+z)×1/2[(x-y)^2+(z-x)^2+(y-z)^2
as (x-y)^2 =x^2 +y^2-2xy
2(x^3+y^3+z^3 -3xyz)=2×1/2 (x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]
2x^3+2y^3+2z^3 -6xyz=(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2] proved
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