$x^2+16x+39$
$\left(198\right)\left(202\right)$
$a^3+b^3-5a^3+6ab^2-2b^3$
$x^4+8x^3+24x^2+32x+16$
$\left(x+6\right)\left(x+3\right)\ge\left(x+1\right)\left(x-1\right)$
$\frac{x-16}{\sqrt[3]{4^2}\cdot\:\sqrt[5]{x}-\frac{\sqrt[3]{x}}{x^{-0,\:2}}}$
$9-10a+1$
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