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\left(3a+b^{2}c^{3}\right)\left(9a^{2}-3ab^{2}c^{3}+b^{4}c^{6}\right)
Rewrite 27a^{3}+b^{6}c^{9} as \left(3a\right)^{3}+\left(b^{2}c^{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).