Find the value of

(i) Here, we see that, $x+y+4=0$

$\therefore \quad x^{3}+y^{3}+(4)^{3}=3 x y(4)$

[using identily, $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$ ]

$=12 x y$   ...(i)

Now, $\quad x^{3}+y^{3}-12 x y+64=x^{3}+y^{3}+64-12 x y$

$=12 x y-12 x y=0$    [from Eq. (i)]

(ii) Here, we see that, $x-2 y-6=0$

$\therefore$  $x^{3}+(-2 y)^{3}+(-6)^{3}=3 x(-2 y)(-6)$

[using identity, $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$ ]

$\Rightarrow$ $x^{3}-8 y^{3}-216=36 x y$ ...(i)

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

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

$=36 x y-36 x y=0$  [from Eq. (i)]