Question

Find the value of $x^3+y^3-12xy+64,$ when $x+y=-4.$

Answer 1

We know that,

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$

If $a+b+c=0$, then $a^3+b^3+c^3=3abc$

Now, given $x^3+y^3-12xy+64$ and

$x+y=-4$

$=>x+y+4=0$

Here, $a=x$, $b=y$, $c=4$ and $a+b+c=x+y+4=0$

Therefore

$x^3+y^3+64=3xy\left(4\right)$

$=12xyz$

Now,
$x^3+y^3+64-12xyz=12xyz-12xyz$
$=0$