Find the value of

$=x^{3}+y^{3}+64-12 x y$

$=x^{3}+y^{3}+4^{3}-3(x)(y)(4)$

$=(x+y+4)\left(x^{2}+y^{2}+4^{2}-x y-y \times 4-4 \times x\right)$

$=(-4+4)\left(x^{2}+y^{2}+16-x y-4 y-4 x\right)$

$[\therefore x+y=-4]=0$

$\therefore x^{3}+y^{3}-12 x y+64=0$