$$( x ^ { 2 } - 4 y ^ { 2 } ) ( x + 3 y ) x = 1 y = 2$$
$x\left(x+6\right)\left(x^{2}-16\right)=2\text{ and }y=2$
$y=2$
$x\left(x+6\right)\left(x^{2}-16\right)=2$
