$x^2-3x-8$
$\int\frac{\sec^2\left(x\right)}{\left(16-\tan^2\left(x\right)\right)^{\frac{3}{2}}}dx$
$3a^3-3ab^2$
$\left(y-5\right)\left(y-8\right)$
$\frac{16x^8-16x^6}{3x-3y}$
$3x^2+6x-1$
$\int x^2\cdot\left(x+1\right)e^xdx$
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