$\frac{\left(x+y\right)^3+\left(x-y\right)^3}{\left(x^2+3y^2\right)}$
$f\left(y\right)=\frac{10}{y^4-2y^2}$
$\int\left(te^{5t\:+\pi}\right)dt$
$27^{3n+6}$
$b^3+6b^2+12b+8$
$10^3\cdot10^5$
$2x-3=f$
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