$3y^2\frac{dy}{dx}=8x$

$z\left(x\right)=-ax^6+bx^3$

$\int tan^5xsec^8xdx$

$\lim_{x\to3}\left(\frac{-4.9x^2+44.1}{x-3}\right)$

$36x^2-9$

$\left(z+m\right)^3+\left(z-m\right)^3=2\left(z^3+6m^3\right)$

$\lim_{x\to3}\left(\frac{\left|x+3\right|}{\left|x\right|-3}\right)$