If $x = 2 y + 6$ , find the value of $x^{3} - 8 y^{3} - 36 x y - 216$

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Solution

It is given that $x = 2 y + 6$ or $x - 2 y = 6$

We know the identity $(a - b)^{3} = a^{3} - b^{3} - 3 a b (a - b)$

Now consider $x - 2 y = 6$ and take cube on both sides and then apply the above identity as follows:

$(x - 2 y)^{3} = (6)^{3} \Rightarrow x^{3} - (2 y)^{3} - 3 x (2 y) (x - 2 y) = 216 \Rightarrow x^{3} - 8 y^{3} - 6 x y (x - 2 y) = 216 \Rightarrow x^{3} - 8 y^{3} - 6 x y (6) = 216 \Rightarrow x^{3} - 8 y^{3} - 36 x y = 216 \Rightarrow x^{3} - 8 y^{3} - 36 x y - 216 = 0$

Hence, $x^{3} - 8 y^{3} - 36 x y - 216 = 0$

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Find the value of:
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(ii) $x^{3} - 8 y^{3} - 36 x y - 216$ , when x = 2y + 6

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The value of $(2 y - x)^{3}$ is _________.

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