$\frac{8\tan^3+1}{8tan^3-4\tan^2+2\tan}$

$\int_1^{-2}3x\left(x-4\right)dx$

$\left(-3\right)^2\cdot\left(-3\right)^0$

$\left(\frac{1}{\cos\left(x\right)}-\frac{\cos\left(x\right)}{1+\sin\left(x\right)}=\tan\left(x\right)sin\right)$

$\left(\frac{3}{5}x+\frac{2}{3}y\right)^2$

$6-5\ln x=8$

$\lim_{x\to-3}\left(\frac{4-\sqrt{x^2-7}}{3x+9}\right)$