$\left(x+3\right)\left(x+6\right)\left(x-4\right)$
$\left(4x^2\:+\:4\right)^2$
$\lim_{x\to0}\left(\frac{\left(e^x-1\right)\tan^2x}{x^3}\right)$
$x^2+\frac{8}{7}+1$
$\sin\left(3x\right)\cos\left(4x\right)+\cos\left(3x\right)\sin\left(4x\right)$
$x+4=28$
$1.1^1$
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