The convert(expr, surd) function converts expressions of the form $x^{\frac{p}{q}}$ to surd(x, q)^p if p and q are integers and $q\ne 1$.

This conversion may not preserve the value of the expression. For example, the exponent is simplified before the conversion is performed.

$x≔{\left(-8\right)}^{\frac{1}{3}}$

$x≔{\left(\mathrm{-8}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$\mathrm{convert}\left(x\,\mathrm{surd}\right)$

$\mathrm{-2}$

$\mathrm{simplify}\left(x\,\mathrm{radical}\right)$

$1+\mathrm{I}\sqrt{3}$

$\mathrm{convert}\left(x^{\frac{1}{3}}{\left(x+y\right)}^{\frac{4}{5}}+{\left(-7\right)}^{\frac{1}{3}}\,\mathrm{surd}\right)$

$-2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(\sqrt[5]{{\left(\mathrm{-8}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+y}\right)}^4-7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$\mathrm{convert}\left(y^{\frac{6}{9}}\,\mathrm{surd}\right)$

${\left(\sqrt[3]{y}\right)}^2$