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\left(2a+3b^{2}\right)\left(4a^{2}-6ab^{2}+9b^{4}\right)
Rewrite 8a^{3}+27b^{6} as \left(2a\right)^{3}+\left(3b^{2}\right)^{3}. The sum of cubes can be factored using the rule: p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right).