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