Find the values:

(A) $x^3 + y^3 - 12xy + 64$ when $x + y = -4$

(B) $x^3 - 8y^3 - 36xy - 216$ when $x = 2y + 6$

(A)  $x^{3} + y^{3} - 12xy + 64$ when $x + y = -4$

$$\text{Rewrite: } x^3 + y^3 + 64 - 12xy$$

$$= x^3 + y^3 + 4^3 - 3xy \cdot 4$$

$$\text{Since, } x^3 + y^3 + 64 - 12xy$$

$$= (x + y + 4)(x^2 + y^2 + 16 - xy - 4y - 4x)$$

$$\text{From } x + y = -4, \text{ we get } x + y + 4 = 0.$$

$$\text{Therefore, } x^3 + y^3 - 12xy + 64 = 0.$$

$$\text{(B) } x^3 - 8y^3 - 36xy - 216 \text{ when } x = 2y + 6$$

$$\text{Rewrite: } x^3 + (-2y)^3 + (-6)^3 - 3x(-2y)(-6)$$

$$= x^3 - 8y^3 - 216 - 36xy$$

$$= (x - 2y - 6)(x^2 + 4y^2 + 36 + 2xy - 12y + 6x)$$

$$\text{From } x = 2y + 6, \text{ we get } x - 2y - 6 = 0.$$

$$\text{Therefore, } x^3 - 8y^3 - 36xy - 216 = 0.$$