Simplify: ${(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3}$

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Solution

We know that $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
or $a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 a b c$
where $a = x - 2 y$ , $b = 2 y - 3 z$ , $z = 3 z - x$
$∴ {(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3}$
$= (x - 2 y + 2 y - 3 z + 3 z - x) ({(x - 2 y)}^{2} + {(2 y - 3 z)}^{2} + {(3 z - x)}^{2} - (x - 2 y) (2 y - 3 z) - (2 y - 3 z) (3 z - x) - (3 z - x) (x - 2 y)) + 3 (x - 2 y) (2 y - 3 z) (3 z - x)$
$= (0) ({(x - 2 y)}^{2} + {(2 y - 3 z)}^{2} + {(3 z - x)}^{2} - (x - 2 y) (2 y - 3 z) - (2 y - 3 z) (3 z - x) - (3 z - x) (x - 2 y)) + 3 (x - 2 y) (2 y - 3 z) (3 z - x)$
$= 0 + 3 (x - 2 y) (2 y - 3 z) (3 z - x)$
$= 3 (x - 2 y) (2 y - 3 z) (3 z - x)$

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