## Intermediate Algebra (12th Edition)

$x^{3}\sqrt[3] x^{2}$
We know from the radical form of $a^{\frac{m}{n}}$ that $a^{\frac{m}{n}}=\sqrt[n] a^{m}$ (where the indicated roots are real numbers). Therefore, $\sqrt[12] x^{44}=x^{\frac{44}{12}}=x^{\frac{11}{3}}=\sqrt[3] x^{11}$. $\sqrt[3] x^{11}=\sqrt[3] (x^{9}\times x^{2})=\sqrt[3] x^{9}\times\sqrt[3] x^{2}=x^{3}\sqrt[3] x^{2}$