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\frac{2\left(27a^{3}+8b^{9}\right)}{81}
Factor out \frac{2}{81}.
\left(3a+2b^{3}\right)\left(9a^{2}-6ab^{3}+4b^{6}\right)
Consider 27a^{3}+8b^{9}. Rewrite 27a^{3}+8b^{9} as \left(3a\right)^{3}+\left(2b^{3}\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).
\frac{2\left(3a+2b^{3}\right)\left(9a^{2}-6ab^{3}+4b^{6}\right)}{81}
Rewrite the complete factored expression.