$a^6-6a^3+9$
$a+\frac{1}{a}=6$
$laplace\:\left(e^t-e^{-t}\right)^2$
$125x^9-64y^{15}$
$\lim_{x\to\infty}\frac{8x^3-3x+4}{5-2x^3}$
$\left(-3\right)^2\left(3\right)$
$\left(\frac{\left(x^5\cdot y^6\right)}{\left(x^3\cdot y^2\right)}\right)^4$
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