$\left(2b-3c\right)^3$
$x+6>4-3x$
$2\cdot\left(\frac{v}{2\pi}\right)^{\frac{1}{3}}$
$54.8+67.27$
$9+6a^2$
$\left(6a^2+6\right)\left(6a^2+2\right)$
$\tan\left(x\right)^2-4=\tan\left(x\right)$
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