Evaluate: (4y-3x)(9x^2+12xy+16y^2)

Factor: $\left(4y-3x\right)\left(9x^{2}+12xy+16y^{2}\right)$

Rewrite $64y^{3}-27x^{3}$ as $\left(-3x\right)^{3}+\left(4y\right)^{3}$. The sum of cubes can be factored using the rule: $a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)$.

$$\left(-3x+4y\right)\left(9x^{2}+12xy+16y^{2}\right)$$