# What is root3(32)/(root3(36))? How do you rationalize the denominator, if needed?

Feb 13, 2016

I got: $2 \frac{\sqrt[3]{81}}{9}$

#### Explanation:

Let us write it as:
$\sqrt[3]{\frac{32}{36}} = \sqrt[3]{\frac{\cancel{4} \cdot 8}{\cancel{4} \cdot 9}} = \frac{\sqrt[3]{8}}{\sqrt[3]{9}} = \frac{2}{\sqrt[3]{9}}$
rationalize:
$= \frac{2}{\sqrt[3]{9}} \cdot \frac{\sqrt[3]{9}}{\sqrt[3]{9}} \cdot \frac{\sqrt[3]{9}}{\sqrt[3]{9}} = 2 \frac{\sqrt[3]{81}}{9}$

Feb 13, 2016

or $\frac{2 \sqrt[3]{3}}{3}$

#### Explanation:

Given $\frac{\sqrt[3]{32}}{\sqrt[3]{36}}$ for rationalizing of denominator if required.
$\sqrt[3]{\frac{32}{36}}$
Dividing the numerator and denominator by common factor 4.
or $\sqrt[3]{{\cancel{32}}^{8} / {\cancel{36}}_{9}}$
or $\sqrt[3]{\frac{8}{9}}$

or 2/root 3((3^2)

[Since $8 = {2}^{3}$, numerator 8 can be written as $\sqrt[3]{{2}^{3}} = 2$.
And denominator 9 can be written as $\sqrt[3]{{3}^{2}}$].

We see that in order to make the exponent of the denominator equal to nearest whole number 1, we need to multiply it by $\sqrt[3]{3}$.

Therefore, multiplying and dividing the numerator and denominator with $\sqrt[3]{3}$
or $2 \cdot \frac{1}{\sqrt[3]{{3}^{2}}} \cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}}$

or $2 \cdot \frac{\sqrt[3]{3}}{3}$