$\left(x+8\right)^2-\left(x-8\right)^2\le-\frac{2}{3}\left(x+49\right)$

$\left(x^3+\:5x^2\:+\:14x\:+\:35\right)\:\:\left(x\:+\:4\right)$

$\sqrt[3]{x}\cdot\sec^2x\frac{-7}{\left(3-x^2\right)}$

$\tan\left(x\right)=8$

$-27x-18$

$\frac{d}{dx}\left(\frac{x^8\left(x-3\right)^4}{\left(x^2+8\right)^8}\right)$

$42+\left(15-2^3\right)$