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\left(-2z^{3}+3r\right)\left(4z^{6}+6rz^{3}+9r^{2}\right)
Rewrite 27r^{3}-8z^{9} as \left(-2z^{3}\right)^{3}+\left(3r\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).