## Precalculus (6th Edition) Blitzer

$x^{6}-4$
This is a difference of squares, $(a+b)(a-b)=a^{2}-b^{2}$ ... $= (x^{3})^{2}-(2)^{2} \quad$... use the rule: $(a^{m})^{n}=a^{mn}$ =$x^{6}-4$