Factor
\frac{\left(2a-3\right)\left(4a^{2}+6a+9\right)}{27}
Evaluate
\frac{8a^{3}}{27}-1
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\frac{8a^{3}-27}{27}
Factor out \frac{1}{27}.
\left(2a-3\right)\left(4a^{2}+6a+9\right)
Consider 8a^{3}-27. Rewrite 8a^{3}-27 as \left(2a\right)^{3}-3^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right).
\frac{\left(2a-3\right)\left(4a^{2}+6a+9\right)}{27}
Rewrite the complete factored expression. Polynomial 4a^{2}+6a+9 is not factored since it does not have any rational roots.
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