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\left(9x^{3}-7a^{4}\right)\left(9x^{3}+7a^{4}\right)
Rewrite 81x^{6}-49a^{8} as \left(9x^{3}\right)^{2}-\left(7a^{4}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).