$\frac{6-x^2}{x}$

$\frac{dy}{dx}\left(x^3y\right)=0$

$\:2x-8+1+5x$

$12\int_0^{\frac{\pi}{2}}\left(-\sin^4\left(x\right)\cos^2\left(x\right)\right)dx$

$\left(\frac{\left(x^2y^{-\frac{1}{6}}z^{\frac{1}{2}}\right)^4}{\left(x^{\frac{1}{3}}y^{\frac{1}{6}}z^{\frac{1}{2}}\right)^{-6}}\right)^{-1}$

$3x^3+9xy^2=9x^3$

$\left(sinx-cosx\right)\left(sinx+cosx\right)=-1+2sin^2x$