$\left(6mn^3+3m^2\right)^2$
$\lim_{x\to1}\left(\frac{x^3+3x^2-9x+5}{2x^2-4x+2}\right)$
$224=-16t^2+144t$
$\left(2x-8\right)-\left(4x+4\right)$
$\frac{\left(x^5+x^2-2\right)}{\left(x+1\right)}$
$\frac{4x^3}{8x^2}$
$100x^2+26x$
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