# How do you rationalize a demoninator if the denominator is a cube root

1. Aug 14, 2013

### mileena

Hi, I know how to rationalize a denominator when it is a square root monomial or a square root binomial (through conjugation).

For example, for a square-root monomial:

5/√25 =

[5(√25)]/
[(√25)(√25)] =

[5(√25)]/25 =

(√25)/5 =

1 or -1

and, for a square-root binomial:

5/(5 + √25) =

5(5 - √25)/
[(5 + √25)(5 - √25)] =

[5(5 - √25)]/
25 -25 =

0/0 [undefined] or 50/0 [undefined]

But what if the denominator is a cube root:

x/(3√11) ?

How do you simplify this (assuming the denominator isn't a perfect cube)? I didn't get a question similar to this correct on my assessment test.

Thanks!

Last edited: Aug 14, 2013
2. Aug 14, 2013

### symbolipoint

The example cube root you ask for seem not tricky because 27 already is a cube. The cubed root of 27 is 3, so the denominator is already rationalized.

How about something like this: $\frac{x}{\sqrt[3]{28}}$
Multiply by 1 in the form of $\frac{\sqrt[3]{28}}{\sqrt[3]{28}}$

Result is $\frac{x\sqrt[3]{28}}{28}$

EDIT: That was already posted in this same posting but already responded to, when I realize now I made a big mistake.

This example is for a cube root. With the (1/(28)^(1/3)) denominator, I should have shown multiplying numerator and denominator this way:
$\frac{\sqrt[3]{28}}{\sqrt[3]{28}}\cdot\frac{\sqrt[3]{28}}{\sqrt[3]{28}}$

Last edited: Aug 14, 2013
3. Aug 14, 2013

### mileena

But doesn't

(3√28) x (3√28) = 3√282) ??

[because 281/3 + 281/3 = 282/3,
because of the rule (xa) (xb) = xa+b ]

Also, I edited my original post to eliminate the perfect cube in the denominator.

Last edited: Aug 14, 2013
4. Aug 14, 2013

### CAF123

Find a value of $n$ in $$\frac{x}{11^{1/3}} \cdot \frac{11^n}{11^n}$$ such that the denominator is rational.

5. Aug 14, 2013

### mileena

That helps a lot!

What if I multiply by: 112/3/112/3

So the denominator will be (111/3)(112/3) = 11 !

So the final answer will be (x3√112)/11

6. Aug 14, 2013

### CAF123

Correct.

7. Aug 14, 2013

### mileena

Thanks! I finally got something right :tongue2:

8. Aug 14, 2013

### symbolipoint

mileena,
please recheck my post. I found my mistake and added better information. We square a square root to bring back the number under the radical. We cube a cubed root to bring back the number under the radical.