Find the cube of $3 x - 2 y - 4 z$ .

x^{3} - y^{3} - 64 z^{3} - 54 x^{2} y + x y^{2} - 108 x^{2} z - 48 y^{2} z + x z^{2} - 96 y z^{2} + x y z

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27 x^{3} - 8 y^{3} - 64 z^{3} - 54 x^{2} y + 36 x y^{2} - 108 x^{2} z - 48 y^{2} z + 144 x z^{2} - 96 y z^{2} + 144 x y z

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27 x^{3} - y^{3} - z^{3} - 54 x^{2} y - x y^{2} - 108 x^{2} z - 48 y^{2} z + 144 x z^{2} - y z^{2} + 144 x y z

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x^{3} - 8 y^{3} - 64 z^{3} - 54 x^{2} y + 36 x y^{2} - 108 x^{2} z - 48 y^{2} z + x z^{2} - 96 y z^{2} - x y z

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Solution

The correct option is B $27 x^{3} - 8 y^{3} - 64 z^{3} - 54 x^{2} y + 36 x y^{2} - 108 x^{2} z - 48 y^{2} z + 144 x z^{2} - 96 y z^{2} + 144 x y z$
Let $3 x - 2 y = a$ , and $4 z = b$
Applying the formula, ${(a - b)}^{3} = a^{3} - b^{3} - 3 a b (a - b)$
$(3 x - 2 y - 4 z)^{3} = (3 x - 2 y)^{3} - (4 z)^{3} - 3 (3 x - 2 y) (4 z) (3 x - 2 y - 4 z) = (3 x)^{3} - (2 y)^{3} - 3 (3 x) (2 y) (3 x - 2 y) - 64 z^{3} - (9 x - 6 y) (12 x z - 8 y z - 16 z^{2}) = 27 x^{3} - 8 y^{3} - 18 x y (3 x - 2 y) - 64 z^{3} - 108 x^{2} z + 72 x y z + 144 x z^{2} + 72 x y z - 48 y^{2} z - 96 y z^{2} = 27 x^{3} - 8 y^{3} - 64 z^{3} - 54 x^{2} y + 36 x y^{2} - 108 x^{2} z + 144 x y z + 144 x z^{2} - 48 y^{2} z - 96 y z^{2}$

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