Without finding the cubes, factorise
$(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$
Without finding the cubes, factorise
$(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$
Let $x-2 y=a, 2 y-3 z=b$ and $3 z-c=x$
$\therefore \quad a+b+c=x-2 y+2 y-3 z+3 z-x=0$
$\Rightarrow \quad a^{3}+b^{3}+c^{3}=3 a b c$
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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