Using the identity and proof: $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ .

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Solution

$x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
First take L.H.S
$(x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
To multiply two polynomials, we multiply each monomial of one polynomial (with its sign) by each monomial (with its sign) of the other polynomial.
$x . x^{2} + x . y^{2} + x . z^{2} - x^{2} y - x y z - x^{2} z + y . x^{2} + y . y^{2} + y . z^{2} - x y^{2} - y^{2} z - x y z + z . x^{2} + z . y^{2} + z . z^{2} - x y z - y z^{2} - x z^{2}$
$=$ $x^{3} + x y^{2} + x z^{2} - x^{2} y - x^{2} y + y x^{2} + y^{3} - x y^{2} - y^{2} z + x^{2} z + y^{2} z + z^{3} - y z^{2} - x z^{2} - 3 x y z$
$=$ $x^{3} + y^{3} + z^{3} - 3 x y z$
L.H.S $=$ R.H.S
$x^{3} + y^{3} + z^{3} - 3 x y z = x^{3} + y^{3} + z^{3} - 3 x y z$
Hence $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ is proved.

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