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

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Solution

Let $a = x - 2 y$ , $b = 2 y - 3 z$ , $c = 3 z - x$ . Then,

$\Rightarrow$ $a + b + c = 0$

$\Rightarrow$ $a^{3} + b^{3} + c^{3} = 3 a b c$

$\Rightarrow$ $(x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3} = 3 (x - 2 y) (2 y - 3 z) (3 z - x)$

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