Factorise the following:
${(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3}$

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Solution

We know the corollary: if $a + b + c = 0$ then $a^{3} + b^{3} + c^{3} = 3 a b c$

Using the above corollary taking $a = (x - 2 y)$ , $b = (2 y - 3 z)$ and $c = (3 z - x)$ , we have $a + b + c = x - 2 y + 2 y - 3 z + 3 z - x = 0$ then the value of $(x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3}$ is:

$(x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3} = 3 [(x - 2 y) \times (2 y - 3 z) \times (3 z - x)] = 3 (x - 2 y) (2 y - 3 z) (3 z - x)$

Hence, $(x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3} = 3 (x - 2 y) (2 y - 3 z) (3 z - x)$

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