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\left(p+3q\right)\left(p^{2}-3pq+9q^{2}\right)
Rewrite p^{3}+27q^{3} as p^{3}+\left(3q\right)^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).