$\left(x^3+6\right)\left(x^4-8\right)$
$\sqrt{2-2cos\left(4x\right)}$
$\:\left(-1\right)-\left(-18\right)$
$\lim_{x\to0}\frac{\sin^3\left(x\right)}{\sin\left(x\right)-\tan\left(x\right)}$
$49x^6-14x^3y^8+y^{16}$
$2.92a+19.03>1.51$
$\left[100\frac{\left(1.05\right)^{12}-1}{0.05}\right]$
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