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\left(3x^{2}+2y^{4}\right)\left(9x^{4}-6x^{2}y^{4}+4y^{8}\right)
Rewrite 27x^{6}+8y^{12} as \left(3x^{2}\right)^{3}+\left(2y^{4}\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).