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\left(5n^{2}+4a\right)\left(25n^{4}-20an^{2}+16a^{2}\right)
Rewrite 64a^{3}+125n^{6} as \left(5n^{2}\right)^{3}+\left(4a\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).