Question

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

Answer 1

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