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