Question

Help! How do I rationalize the denominator in this problem : 3rd root of 1/6 , so that the result is in simplest form?

Answer 1

= 6 ^(2/3) / 6

Answer 2

How do you get 6^(2/3) out of
\[\sqrt[3]{1/6}\]

Answer 3

I get 3rd root of 6/6

Answer 4

The idea here is to break it apart like this since \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \).
\[ \sqrt[3]{\frac{1}{6}} = \frac{\sqrt[3]{1}}{\sqrt[3]{6}} \]

To rationalize this, we need to multiply the numerator and the denominator by a factor that cancels the cube root in the denominator. We know that the number on the inside has to be a \(\textbf{perfect cube}\) to simplify, so we have to think about what we can multiply to \(\sqrt[3]{6}\) to create that perfect cube, and then multiply it to both the numerator and the denominator..

Answer 5

would I multiply the numerator and denominator by
\[\sqrt[3]{6}?\]

Answer 6

that was my first thought when I wrote the problem down, to multiply by \[\sqrt[3]{6}\]

Answer 7

Well, what happens then, is that we get this:
\[ \frac{\sqrt[3]{6}}{\sqrt[3]{6^2}} = \frac{\sqrt[3]{6}}{\color{green}{\sqrt[3]{36}}}\]

But is 36 a perfect cube that we can take out? Or maybe we need to multiply by another \(\sqrt[3]{6}\) to make it a perfect cube? :)

Answer 8

If we multiply by another cube root of 6 to both numerator and denominator, we create a \(\large \sqrt[3]{6^3}\) in the denominator, so the cube root and the cube cancel to leave 6.
\[ \frac{\sqrt[3]{36}}{6} \]

That numerator is also 6^2/3, as previously mentioned.

Answer 9

oh,
when I said earlier,
"would I multiply the numerator & denominator by \[\sqrt[3]{6}\] " ...isn't that what you just said now? "to multiply by another cube root of 6 to both numerator and denominator"?

Answer 10

The idea is to make a perfect cube under the cube root with the denominator.
When we multiplied by the first cubed root of 6, we got cuberoot(36) in the denominator, which isn't a perfect cube. We had to instead multiply by another cuberoot(6) to make it a perfect cube. (We multiply 6*6*6, 6^3, so that's a perfect cube).

Effectively, we multiplied the numerator and the denominator by \(\sqrt[3]{6^2}\) , or 6^(2/3), since we had 6^(1/3) already.

\[ \frac{1}{\sqrt[3]{6}} \times \frac{\sqrt[3]{6^2}}{\sqrt[3]{6^2}} \]

Answer 11

Unlike square roots, which you just need one other square root to make it 'complete':
\(\sqrt{2} \times \sqrt{2} = 2\)

Cube roots need three:
\(\sqrt[3]{2} \times \sqrt[3]{2} \times \sqrt[3]{2} = 2\)

Answer 12

ok, the denominator would be 6, and the numerator would be \[\sqrt[3]{36}\]?

Answer 13

Yeah. That's correct.

Answer 14

I think I see that now, if the denominator is, for example, \[\sqrt[3]{2}\],
you would multiply numerator and denominator by
\[\sqrt[3]{2^{2}}\] ?

Answer 15

Yeah. We complete the cube under the cube root and it pretty much just cancels with the root to get the whole number.

Answer 16

What if the denominator is for example,

\[\sqrt[5]{8}\]

would you multiply numerator and denominator by

\[\sqrt[5]{8^{4}}\] ?

Answer 17

Yep. Or, if you think about it, \(8=2^3\), so we could also multiply by \(\sqrt[5]{2^2}\) since that makes \(\sqrt[5]{2^5}\).
If you do it your way, usually you just have to do some extra simplifying and you'll get the simplest form.

Answer 18

Little tricks like that come up every so often when you have some perfect squares, cubes, etc. as factors.

Answer 19

The key is we want exponent is 1
=> 1 - 1/5 = 4/5

Answer 20

By fraction rule, if we multiply/ divide => it need to be done on BOTH upper and lower!

Answer 21

@AccessDenied you always deserve multi medals more than I'm ;)

Answer 22

*deserved

Answer 23

Thank you for your help, AccessDenied & Chlorophyll!!...both of you have made this type of problem easier to understand1

Answer 24

I just copy cat the idea after @AccessDenied :P

Answer 25

when something is raised to a fractional exponent, it makes the problem seem difficult, at first! ...thanks again for your help!

Answer 26

You're welcome! The rational exponents, I find, are usually easier to represent some of the properties like that previous cancelling the square and sixth root.
It's just harder to imagine a power as a fraction since it usually means "take this multiplied to itself # times," but as a fraction, its like "take this multiplied to itself but only half-way". The fraction just indicates that we don't have the 'whole' thing there.