Question

Prove that x3 y3 z3 - 3xyz = [x y z] [x2 y2 z2 - xy -yz - zx]

Answer 1

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Answer 2

Answer:

$x^{3}+y^{3}+z^{3}-3xyz\\=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

Step-by-step explanation:

$Proving\: algebraic\: identity:\\x^{3}+y^{3}+z^{3}-3xyz\\=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

$RHS=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

$=x(x^{2}+y^{2}+z^{2}-xy-yz-zx)\\+y(x^{2}+y^{2}+z^{2}-xy-yz-zx)\\+\\z(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

$=x^{3}+xy^{2}+xz^{2}-x^{2}y-xyz-zx^{2}\\+x^{2}y+y^{3}+yz^{2}-xy^{2}-y^{2}z-xyz\\+x^{2}z+y^{2}z+z^{3}-xyz-yz^{2}-z^{2}x$

$=x^{3}+y^{3}+z^{3}-3xyz\\=LHS$

Therefore,.

$x^{3}+y^{3}+z^{3}-3xyz\\=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

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