$\left(2x-5y\right)\left(2x+8y\right)$
$2x-8=x+1$
$\frac{x^3-8}{2x^2}$
$8\:-\:\left(-2\right)\:+\:3\:-\:5$
$x^2-3y^2-2z^2-9=0$
$0\cdot5+5$
$-\frac{12}{8}\left(\frac{t^{12}}{12}\right)$
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