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\left(xy+4z\right)\left(x^{2}y^{2}-4xyz+16z^{2}\right)
Rewrite x^{3}y^{3}+64z^{3} as \left(xy\right)^{3}+\left(4z\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).