Question

Find the value of x³-8y³-36xy-216, when x=2y+6

Answer 1

Here it is! Hope it helps ^-^

Answer 2

It is given that $x = 2y+6$. 

$x^{3} - 8y^{3} - 36xy -216$
$= \left(2y+6\right)^3-8\left(2y+6\right)^3-36\left(2y+6\right)\left(y\right)-216$
$=\left(2y+6\right)^3-8\left(2y+6\right)^3-36\left(2y+6\right)y-216$
$=-7\left(2y+6\right)^3-36y\left(2y+6\right)-216$
$=-7\left(8y^3+72y^2+216y+216\right)-36\left(2y+6\right)y-216$
$=-56y^3-504y^2-1512y-1512-36\left(2y+6\right)y-216$
$=-56y^3-504y^2-1512y-1512-72y^2-216y-216$
$=-56y^3-576y^2-1728y-1728$

Hope this may help you.