$\frac{\left(x^3-x\right)}{x^2-1}$
$\sec\left(x\right)^2+\tan^2\left(x\right)=1$
$\int\left(\frac{3x^2-x+8}{x^3+4x}\right)dx$
$\left(7a^2+2b^3-8x\right)^2$
$x^2+24x=12$
$\left(4u\:\:-\:\:5v\right)^2$
$y2\:+\:8\:=\:6y$
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