The factors of $(2 x - 3 y)^{3} + (3 y - 4 z)^{3} + 8 (2 z - x)^{3}$ are :

2 x \times 3 y \times 4 z

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2 (2 x - 3 y) (3 y - 4 z) (2 z - x)

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6 (2 x - 3 y) (3 y - 4 z) (2 z - x)

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3 (2 x - 3 y) (3 y - 4 z) (2 z - x)

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Solution

The correct options are
B $2 (2 x - 3 y) (3 y - 4 z) (2 z - x)$
C $6 (2 x - 3 y) (3 y - 4 z) (2 z - x)$
D $3 (2 x - 3 y) (3 y - 4 z) (2 z - x)$
$({2 x - 3 y)}^{3} + {(3 y - 4 z)}^{3} + {(4 z - 2 x)}^{3}$
The equation can be written in this way
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 $({a)}^{3} + {(b)}^{3} + {(c)}^{3} = 3 a b c$
Hence as we see, $2 x - 3 y + 3 y - 4 z + 4 z - 2 x = 0$
$({2 x - 3 y)}^{3} + {(3 y - 4 z)}^{3} + {(4 z - 2 x)}^{3}$
$= 6 (2 x - 3 y) (3 y - 4 z) (2 z - x)$
hence, option B,C,D are correct.

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{(2 x - 3 y)}^{3} + {(4 z - 2 x)}^{3} + {(3 y - 4 z)}^{3}

$(2 x - 3 y)^{3} + (x + 3 y)^{3} + 3 (2 x - 3 y)^{2} (x + 3 y) + 3 (2 x - 3 y) (x + 3 y)^{2} =$

(2 x - 3 y)^{3} + (x + 3 y)^{3} + 3 (2 x - 3 y)^{2} (x + 3 y) + 3 (2 x - 3 y) (x + 3 y)^{2} =

(2 x - 3 y)^{3} + (x + 3 y)^{3} + 3 (2 x - 3 y)^{2} (x + 3 y) + 3 (2 x - 3 y) (x + 3 y)^{2}

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