Factorise the following : $27 (a - b)^{3} + (2 a - b)^{3} + (4 b - 5 a)^{3}$

(a - b) (a - b) (4 b - 8 a)

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9 (a - b) (2 a - b) (4 b - 5 a)

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9 (a - b) (a - 2 b) (4 b - a)

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94 - b) (2 a - b) (b - a)

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Solution

The correct option is B $9 (a - b) (2 a - b) (4 b - 5 a)$
We know,
$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
If, $a + b + c = 0$ then,
$a^{3} + b^{3} + c^{3} - 3 a b c = 0$
$=> a^{3} + b^{3} + c^{3} = 3 a b c$
Now,
$27 (a - b)^{3} + (2 a - b)^{3} + (4 b - 5 a)^{3}$
$= (3 a - 3 b)^{3} + (2 a - b)^{3} + (4 b - 5 a)^{3}$
Here,
$a = 3 a - 3 b$
$b = 2 a - b$
$c = 4 b - 5 a$
Now,
$a + b + c = 3 a - 3 b + 2 a - b + 4 b - 5 a$
$= 0$
Thus,
$(3 a - 3 b)^{3} + (2 a - b)^{3} + (4 b - 5 a)^{3} = 9 (a - b) (2 a - b) (4 b - 5 a)$

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