$x=8x$
$\frac{\left(\sqrt{x^3-3x}\cdot\sqrt[3]{e^x}\right)}{x^2}$
$\sqrt[3]{x^2}\sqrt[5]{y^3}\sqrt[7]{z^5}$
$\left(2c^3+b\right)^2$
$-7y+3y$
$\left(\frac{2}{3}x^2y-\frac{1}{4}\right)\left(\frac{2}{3}x^2+\frac{1}{4}\right)$
$7n+4n$
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