$\left(\frac{x-3}{x^2-25}\right)-\frac{\left(x-5\right)}{\left(3-x\right)}$
$\left(9+3\sqrt[3]{4}+\sqrt[3]{16}\right)$
$c^4.c^8$
$\left(4wz^2-4w^2z\:\right)^2$
$6x+4=9x-5$
$6a^2+22a+20$
$2\le-4+2p$
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