# How do I rationalize Cubic root?

Apr 2, 2015

Use the fact that $\sqrt[3]{{a}^{3}} = a$
Together with $\sqrt[3]{a} \sqrt[3]{b} = \sqrt[3]{a b}$
(And, of course, the fact that multiplying by one may change the way a number is written, but it does not change the (value of) the number.)
Example 1
Rationalize the denominator: $\frac{2}{\sqrt[3]{5}}$

We'll use the facts mentioned above to write:

$\frac{2}{\sqrt[3]{5}} = \frac{2}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{{5}^{2}}}{\sqrt[3]{{5}^{2}}} = \frac{2 \sqrt[3]{25}}{\sqrt[3]{{5}^{3}}} = \frac{2 \sqrt[3]{25}}{5}$

Example 2
Rationalize the denominator: $\frac{7}{\sqrt[3]{4}}$ .
We could multiply by $\frac{\sqrt[3]{{4}^{2}}}{\sqrt[3]{{4}^{2}}}$, but $\sqrt[3]{16}$ is reducible!

We'll take a more direct path to the solution if we Realize that what we have is:$\frac{7}{\sqrt[3]{{2}^{2}}}$ so we only need to multiply by $\frac{\sqrt[3]{2}}{\sqrt[3]{2}}$,

$\frac{7}{\sqrt[3]{4}} = \frac{7}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \frac{7 \sqrt[3]{2}}{\sqrt[3]{{2}^{3}}} = \frac{7 \sqrt[3]{2}}{2}$

Example 3 (last)
If the denominator is $\sqrt[3]{20}$, the similar path to rationalizing would be:

$\sqrt[3]{20} = \sqrt[3]{{2}^{2} \cdot 5}$, so we would multiply by

think . . . (what is missing to make a perfect cube?

think some more, you can get it. . .

.

.

multiply numerator and denominator by $\sqrt[3]{2 \cdot {5}^{2}}$.