$\left(\frac{2}{3}x^3-\frac{1}{3}x^2\right)^2$
$( x ^ { 5 } ) ^ { \frac { 1 } { 8 } }$
$\left(\frac{x}{4}+4\right)\left(\frac{x}{4}-6\right)$
$6\left(4-x\right)=0$
$\frac{2x+2}{8x+3}$
$x^2+bx+\frac{b}{2}^2$
$b^6\:\cdot\:b^4$
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