$\left(x^3-3x^2+4x-4\right)\cdot\left(3x^2-3x+5\right)$
$y'=\frac{y+1}{x-1}$
$\left(q^{-4}\right)^{-9}$
$5x^6+8x^4+\frac{1}{8}x^3$
$\int\frac{5x^2-28x+20}{5x-3}dx$
$\frac{\left(12x^3+16x^2+9x+2\right)}{\left(6x-1\right)}$
$-6-4\cdot8+11\cdot2$
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