Question

Explain why multiplying $\frac{7}{\sqrt[3]{x}}$by$\frac{\sqrt[3]{x}}{\sqrt[3]{x}}$

does not rationalize the denominator.

Short answer

In order to rationalize the denominator, multiply numerator and denominator by an expression such that denominator results into no radicals. In the given expression $\frac{7}{\sqrt[3]{x}}$,the denominator $\sqrt[3]{x}$has only one power of $x$.Hence, multiply numerator and denominator by $\sqrt[3]{x^2}$to make the perfect cube. Thus, multiplying the given expression $\frac{7}{\sqrt[3]{x}}$by $\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}$results in rationalization.

Step-by-step solution

Step 1 The given expression is,

$\frac{7}{\sqrt[3]{x}}$.

The objective is, we need to give the reason for multiplying the expression $\frac{7}{\sqrt[3]{x}}$by $\frac{\sqrt[3]{x}}{\sqrt[3]{x}}$does not result in rationalization.

Step 2 Reason for multiplying the given expression by x3x3 does not result in rationalization :

In order to rationalize the denominator, multiply numerator and denominator by an expression such that denominator results into no radicals. In the given expression $\frac{7}{\sqrt[3]{x}}$, the denominator $\sqrt[3]{x}$has only one power of $x$. Hence, multiply numerator and denominator by $\sqrt[3]{x^2}$to make the perfect cube.

Thus, multiplying the given expression $\frac{7}{\sqrt[3]{x}}$by $\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}$results in rationalization.

Therefore, the reason for multiplying the expression $\frac{7}{\sqrt[3]{x}}$by $\frac{\sqrt[3]{x}}{\sqrt[3]{x}}$does not result in rationalization is explained.