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\left(ba^{6}-3c\right)\left(b^{2}a^{12}+3bca^{6}+9c^{2}\right)
Rewrite a^{18}b^{3}-27c^{3} as \left(ba^{6}\right)^{3}-\left(3c\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).