$x+5-4x-8$
$\frac{1}{3}x^3+\frac{2}{3}x^2-3x-6$
$-3\:x\:4-9$
$6^5\cdot6^2$
$2.\left(x-3\right)-\left(2x^3+\frac{2}{3}x\right):\left(x-3\right)$
$-5\:\left(-4\right)\:\left(-25\right)$
$x^4+17x^2-60$
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