$\frac{3}{4}\sin\left(x\right)-\frac{1}{4}\sin\left(3x\right)=\sin^3\left(x\right)$
$3\:\le8+x$
$\left(-3m^9\right)\left(4m^{-5}\right)$
$1\cdot3.1416$
$-7\left(-1-n\right)$
$\left(k+6\right)\left(k-\frac{2}{3}\right)$
$2\ln\left(x\right)+\ln\left(x^4+2x^2\right)$
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