Factorise : ${(5 x - y)}^{3} + {(y - 4 z)}^{3} + {(4 z - 5 x)}^{3}$

{(5 x - y)}^{3} + {(y - 4 z)}^{3} + {(z - x)}^{3} = 3 (x - y) (y - 4 z) (4 z - 5 x)

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{(5 x - y)}^{3} + {(y - 4 z)}^{3} + {(4 z - 5 x)}^{3} = 3 (5 x - y) (y - 4 z) (4 z - 5 x)

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{(5 x - y)}^{3} + {(y - 4 z)}^{3} + {(z + 5 x)}^{3} = 3 (5 x - y) (y - 4 z) (4 z - 5 x)

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{(x - y)}^{3} + {(y - 4 z)}^{3} + {(4 z - 5 x)}^{3} = 3 (5 x - y) (y - 4 z) (z - 5 x)

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Solution

The correct option is B ${(5 x - y)}^{3} + {(y - 4 z)}^{3} + {(4 z - 5 x)}^{3} = 3 (5 x - y) (y - 4 z) (4 z - 5 x)$
Let $a = 5 x - y$ , $b = y - 4 z$ , $c = 4 z - 5 x$
$a + b + c = 5 x - y + y - 4 z + 4 z - 5 x = 0$
$∴ a^{3} + b^{3} + c^{3} = 3 a b c$
$\Rightarrow {(5 x - y)}^{3} + {(y - 4 z)}^{3} + {(4 z - 5 x)}^{3} = 3 (5 x - y) (y - 4 z) (4 z - 5 x)$

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