Factorise the following : $(5 x - 6 y)^{3} + (7 z - 5 x)^{3} + (6 y - 7 z)^{3}$

3 (5 x - 6 y) (7 z - 5 x) (6 y - 7 z)

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(x - 6 y) (7 z - x) (y - 7 z)

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3 (x - 6 y) (z + 5 x) (8 y - z)

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(x - y) (7 z + 5 x) (6 y - 7 z)

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Solution

The correct option is A $3 (5 x - 6 y) (7 z - 5 x) (6 y - 7 z)$
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,
$(5 x - 6 y)^{3} + (7 z - 5 x)^{3} + (6 y - 7 z)^{3}$
Here,
$a = 5 x - 6 y$
$b = 7 z - 5 x$
$c = 6 y - 7 z$
Now,
$a + b + c = 5 x - 6 y + 7 z - 5 x + 6 y - 7 z$
$= 0$
Thus,
$(5 x - 6 y)^{3} + (7 z - 5 x)^{3} + (6 y - 7 z)^{3} = 3 (5 x - 6 y) (7 z - 5 x) (6 y - 7 z)$

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