Question 38
Without finding the cubes, factorise:
$(x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3}$

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Solution

We know that,

$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

Also, if a + b + c = 0,
then $a^{3} + b^{3} + c^{3} = 3 a b c$ .....condition(1)

Here, we see that
$a + b + c = (x - 2 y) + (2 y - 3 z) + (3 z - x) = 0$

$∴ Using condition (1), we get \Rightarrow (x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3}$
$= 3 (x - 2 y) (2 y - 3 z) (3 z - x)$

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