Factorise: $27 x^{3} + y^{3} + z^{3} - 9 x y z$

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Solution

It is known that, $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)$
$∴ 27 x^{3} + y^{3} + z^{3} - 9 x y z = (3 x)^{3} + (y)^{3} + (z)^{3} - 3 (3 x) (y) (z)$
$= (3 x + y + z) [(3 x)^{2} + (y)^{2} + (z)^{2} - (3 x) (y) - (y) (z) - (3 x) (z)]$
$= (3 x + y + z) [9 x^{2} + y^{2} + z^{2} - 3 x y - y z - 3 x z]$
Hence, $27 x^{3} + y^{3} + z^{3} - 9 x y z = (3 x + y + z) [9 x^{2} + y^{2} + z^{2} - 3 x y - y z - 3 x z]$

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