$\left(\frac{y^5}{3x^3}\right)-4$
$\left(mn^2+4\right)^2$
$\left(14a^2+a\right)^2$
$\left(x+5x^2\right)^3$
$5x^3+15x^2+5$
$\tan\left(v\right)^5=\left(\sec\left(v\right)^4-2\sec\left(v\right)^2+1\right)\tan\left(v\right)$
$\frac{10x}{10x^2}$
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