$\left(4+\frac{1}{3}x\right)^4$
$x\frac{dy}{dx}=x^3+3y$
$\left(b^{-2}\right)^5$
$\frac{6x^4-2x^2+5x-3}{x-2}$
$\left(-2x^2-x\right)^3$
$\frac{6x^2-4^2+7x+3}{3x+1}$
$s^2+2s-8$
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