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

Solution :
We have
\begin{aligned} & x^{3}-8 y^{3}-36 x y-216 \\ =& x^{3}+\left(-8 y^{3}\right)+(-216)-36 x y \\ =& x^{3}+(-2 y)^{3}+(-6)^{3}-3 \times x \times(-2 y) \times(-6) \\ =& a^{3}+b^{3}+c^{3}-3 a b c, \text { where } a=x, b=-2 y \text { and } c=-6 \\ =&(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\ =&(x-2 y-6)\left(x^{2}+4 y^{2}+36+2 x y-12 y+6 x\right) \\ =& 0 \times\left(x^{2}+4 y^{2}+36+2 x y-12 y+6 x\right)=0 \end{aligned}

\begin{aligned} & x^{3}-8 y^{3}-36 x y-216 \\=& x^{3}+\left(-8 y^{3}\right)+(-216)-36 x y \\=& x^{3}+(-2 y)^{3}+(-6)^{3}-3 \times x \times(-2 y) \times(-6) \\=& a^{3}+b^{3}+c^{3}-3 a b c, \text { where } a=x, b=-2 y \text { and } c=-6 \\=&(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\=&(x-2 y-6)\left(x^{2}+4 y^{2}+36+2 x y-12 y+6 x\right) \\=& 0 \times\left(x^{2}+4 y^{2}+36+2 x y-12 y+6 x\right)=0 \\=&[\because x-2 y-6=0 \text { (given)]. }\end{aligned}

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

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