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

[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 -1and, 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!

 

[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}}

 

[mileena]

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}

But doesn't

(³√28) x (³√28) = ³√28²) ??

[because 28^1/3 + 28^1/3 = 28^2/3,
because of the rule (x^a) (x^b) = x^a+b ]

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

 

[CAF123]

But doesn't

(³√28) x (³√28) = ³√28²) ??

(because the exponents 1/3 + 1/3 = 2/3)

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

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

 

[mileena]

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

That helps a lot!

What if I multiply by: 11^2/3/11^2/3

So the denominator will be (11^1/3)(11^2/3) = 11 !

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

 

[CAF123]

That helps a lot!

What if I multiply by: 11^2/3/11^2/3

So the denominator will be (11^1/3)(11^2/3) = 11 !

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

Correct.

 

[mileena]

Thanks! I finally got something right

 

[symbolipoint]

Thanks! I finally got something right

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.