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\left(3x-4y\right)\left(9x^{2}+12xy+16y^{2}\right)
Rewrite 27x^{3}-64y^{3} as \left(3x\right)^{3}-\left(4y\right)^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).