The textbook context provides the algebraic identity $(x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) = x^3 + y^3 + z^3 - 3xyz$. By setting $x + y + z = 0$, we can derive that $x^3 + y^3 + z^3 = 3xyz$. For part (i), substituting the terms $x, y,$ and $4$ allows us to evaluate the expression since their sum is $x+y+4=0$. For part (ii), substituting $x, -2y,$ and $-6$ works perfectly because their sum is $x-2y-6=0$.