$4+4x+4^2$
$\left(tan^2x-sec^2x\right)-1$
$\frac{x^3+5x+3}{x-4}$
$-\frac{8}{3}\:and\:-1$
$\left(x^5-3ay^2\right)^2$
$\frac{\left(+55\right)}{\left(+5\right)}$
$4^{-2}\cdot4^4$
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