$\frac{x^3+3x^2-6x-8}{x+4}=0$
$2021x^2+32x-1\cdot6x^3-6x^2-7x-6$
$5x^3-6x^2+2x-1$
$a3-ab\cdot a$
$81x^2-72x+17$
$\frac{\left(3x^3-10x^2-7x-5\right)}{\left(3x-4\right)}$
$\frac{d}{dx}+2xy=x^3$
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