$\int\frac{16}{x^3-4x^2}dx$
$\sin\left(16x\right)=16\sin\left(x\right)\cos\left(x\right)\cos\left(2x\right)\cos\left(4x\right)\cos\left(8x\right)$
$4b\left(1\cdot7\right)$
$\lim_{x\to0}\left(\frac{sin\left(4x\right)-x^2}{x}\right)$
$g^{-3}h$
$2\cdot10^{5}$
$43\cdot43\cdot43$
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