Given:

\(\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}}\)

“For any rational exponent min in lowest terms, where m and n are integers and \(n\ >\ 0,\) we define

\(a^{^m/_n}=(\sqrt[n]{a})^{m}=\sqrt[n]{a^{m}}\)

If nis even, then we require that \(a\ \geq\ 0”\)

Calculation:

Consider the given expression,

\(\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}}\)

Apply radical rule : \(\sqrt[n]{ab}=\sqrt[n]{a}\ \sqrt[n]{b}\) in the numerator, we get

\(\sqrt[3]{8x^{2}}=\sqrt[3]{8}\ \cdot\ \sqrt[3]{x^{2}}\)

\(=\sqrt[3]{2^{3}}\ \cdot\ \sqrt[3]{x^{2}}\)

\(=2\sqrt[3]{x^{2}}\)

Substitute \(\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}}=2\sqrt[3]{x^{2}}\) we get

By using the law of exponents,

\(\frac{2\sqrt[3]{x^{2}}}{x}=\frac{2x^{\frac{2}{3}}}{x^{\frac{1}{2}}}\)

Apply exponent rule: \(\frac{x^{a}}{x^{b}}=x^{a\ -\ b}\) we get

\(\frac{2x^{\frac{2}{3}}}{x^{\frac{1}{3}}}=2x^{\frac{2}{3}\ -\ \frac{1}{2}}\)

\(= 2x^{\frac{1}{6}}\)

Final statement:

The simplyfied form of \(\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}} is 2x^{\frac{1}{6}}\)

\(\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}}\)

“For any rational exponent min in lowest terms, where m and n are integers and \(n\ >\ 0,\) we define

\(a^{^m/_n}=(\sqrt[n]{a})^{m}=\sqrt[n]{a^{m}}\)

If nis even, then we require that \(a\ \geq\ 0”\)

Calculation:

Consider the given expression,

\(\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}}\)

Apply radical rule : \(\sqrt[n]{ab}=\sqrt[n]{a}\ \sqrt[n]{b}\) in the numerator, we get

\(\sqrt[3]{8x^{2}}=\sqrt[3]{8}\ \cdot\ \sqrt[3]{x^{2}}\)

\(=\sqrt[3]{2^{3}}\ \cdot\ \sqrt[3]{x^{2}}\)

\(=2\sqrt[3]{x^{2}}\)

Substitute \(\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}}=2\sqrt[3]{x^{2}}\) we get

By using the law of exponents,

\(\frac{2\sqrt[3]{x^{2}}}{x}=\frac{2x^{\frac{2}{3}}}{x^{\frac{1}{2}}}\)

Apply exponent rule: \(\frac{x^{a}}{x^{b}}=x^{a\ -\ b}\) we get

\(\frac{2x^{\frac{2}{3}}}{x^{\frac{1}{3}}}=2x^{\frac{2}{3}\ -\ \frac{1}{2}}\)

\(= 2x^{\frac{1}{6}}\)

Final statement:

The simplyfied form of \(\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}} is 2x^{\frac{1}{6}}\)