$\left(8x^2-3y^6\right)^3$
$\frac{8}{3}-n=\frac{3}{7}$
$\left(\frac{x^2}{y^2}\right)^{-1}$
$25y^6-10y^3+1$
$-2\left(-4\right)\left(-1\right)\left(-4\right)$
$x+5=-14$
$\left(1^5\right)^3$
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