# Simplify the expression and express the answer using rational exponents. Assume that all letters denote positive numbers. frac{sqrt[3]{8x^{2}}}{sqrt{x}}

Simplify the expression and express the answer using rational exponents. Assume that all letters denote positive numbers.
$$\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}}$$

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Tasneem Almond
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 $$\frac{\sqrt[3]{8x^{2}}}{\sqrt{x}} is 2x^{\frac{1}{6}}$$