Factorise the following:
${(2 x - 3 y)}^{3} + {(4 z - 2 x)}^{3} + {(3 y - 4 z)}^{3}$

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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 $a = (2 x - 3 y)$ , $b = (4 z - 2 x)$ and $c = (3 y - 4 z)$ , we have $a + b + c = 2 x - 3 y + 4 z - 2 x + 3 y - 4 z = 0$ then the value of $(2 x - 3 y)^{3} + (4 z - 2 x)^{3} + (3 y - 4 z)^{3}$ is:

$(2 x - 3 y)^{3} + (4 z - 2 x)^{3} + (3 y - 4 z)^{3} = 3 [(2 x - 3 y) \times (4 z - 2 x) \times (3 y - 4 z)] = 3 (2 x - 3 y) (4 z - 2 x) (3 y - 4 z)$
$= 6 (2 x - 3 y) (2 z - x) (3 y - 4 z)$

Hence, $(2 x - 3 y)^{3} + (4 z - 2 x)^{3} + (3 y - 4 z)^{3} = 6 (2 x - 3 y) (2 z - x) (3 y - 4 z)$

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