$\left(-26\right)\left(23\right)$
$\lim_{x\to\infty}\left(\frac{\left(2x^2-6x-8\right)}{6}\right)$
$\left(2p+q\right)\left(2p-q\right)\left(4p^2+q^2\right)$
$\left(2x^3-3x+2\right):\left(2x-1\right)$
$ln\left(x-4\right)+\frac{1}{x-4}$
$-xy-1-3x=0$
$x^2y+2xy^2+y^3$
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