$\frac{\sin2x+\sin6x}{\cos2x+\cos6x}=\tan\left(4x\right)$
$\left(m^{-2}\right)^2$
$\left(6a^4-b^5\right)^3$
$\:-\:2\:x\:\le\:6\:$
$-10-\left(-7\right)+\left(-9\right)$
$\int\left(2x^2+x\frac{1}{2}-3e^x\right)dx$
$4m^2+16b^2+24ab$
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