$2\left(3x-2\right)^4$
$8\sin^3\left(x\right)+1=6sin\left(x\right)$
$\lim_{x\to\infty}\left(\sqrt{\left(2x^2+1\right)}-2x\right)$
$2-\left|-5-2\left(-1-2+6-1\right)\right|-4\cdot\left(-3+5\right)$
$\frac{\left(-3x^3+x^2+1\right)}{\left(x^2+3\right)}$
$24n^2-3n^2$
$a\left(x+4\right)+b\left(x+4\right)+c\left(x+4\right)$
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