$\lim_{x\to\infty}\left(\frac{1}{6x+3}\right)$
$-x^3-3x^2-3x-3$
$-3\left(-4\right)+\left(2-2^2\right)^3$
$\frac{2x^3-5x^2+6x-3}{x+2}$
$\int\left(2\cdot\left(2x+6\right)^3\right)dx$
$\left(+63\right):\left(-21\right)$
$\frac{d}{dx}x^3+3x^2y+y^3=8$
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