$x\left(x+24\right)$
$\left(6a\right)^4$
$y^3+3y^2-4y=0$
$\frac{5}{4}x\left(\frac{1}{2}\right)$
$\frac{6x^3-3x^2-21x-11}{3x+3}$
$\frac{1+tan\left(\frac{x}{2}\right)^2}{1-tan\left(\frac{x}{2}\right)^2}$
$-7u+5\left(u-3\right)=-1$
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