Help please?

Question

Rationalize this problem: cube root of 3a^2 over cube root of 9a

Answer 1

It's the same thing as cube root (3a^2 / 9a) which simplifies to:
cube root (a / 3).

You don't want a radical in the denominator, so multiply top and bottom by 9.
cube root (9a / 27)

This comes out to cube root (9a) / cube root (27) or:
cube root (9a) / 3.

Answer 2

Cubed root can also be written as ^(1/3) so that's what I will do to make it easier to read.

In order to rationalize a fraction with a root on the denominator, you must multiply the numerator and the denominator by a number that will make the denominators root dissapear.

In this case you don't need to multiply a number because if you distribute the cubed roots, you will see that it cancells out itself.

(3a^2)^(1/3) / (9a)^(1/3)
= 3^(1/3) * (a^2)^(1/3) / 9^(1/3) * a^(1/3)
= 3^(1/3) * a^(2/3) / 9^(1/3) * a^(1/3)
= 3^(1/3) * a^(1/3) / 9^(1/3)

Since x^a / x^b = x^(a-b)

This gets rid of the cubed root variable in the denominator.
BUT you can simplify this further.

We notice that 9 = 3^2
So...

3^(1/3) * a^(1/3) / 9^(1/3)
= 3^(1/3) * a^(1/3) / (3^2)^(1/3)
= 3^(1/3) * a^(1/3) / 3^(2/3)
= 3^(-1/3) * a^(1/3)
= a^(1/3) / 3^(1/3)

Therefore, the simplified answer is:

cubed root of a over cubed root of 3

Answer 3

In general, to rationalize a denominator, you want to give it more stuff so that the radical in the denominator becomes "do-able".

Your denominator is cuberoot(9a). As puzzling pointed out, you could simplify your fraction first, but that kind of trivializes this technique in general. So although his approach (her?) is best for this particular problem, allow me to generalize. Here's how to rationalize ANY denominator.

1) Factor the radicand (part inside the radical) COMPLETELY, using exponents where applicable.

Ex: fourthroot(8 a^2 b^5 c^8) factors into
fourthroot( 2^3 a^2 b^5 c^8)

2) Your next objective is to bump up each power until it is divisible by the index (the index of fourthroot is 4). Do this by multiplying enough factors under the same radical sign.

Ex: fourthroot( 2^3 a^2 b^5 c^8)

The exponent on the 2 is currently 3. The next number after 3 that is divisble by four is 4. So I need one more 2 in there.

Similarly, I need two more a's, three more b's, and no more c's.

So the radical I would use to rationalize the denominator is fourthroot(2 a^2 b^3).




Once you determine what radical will work, use it on the top and bottom of your fraction.


Particular to your denominator, you would do the following:

1) cuberoot(9a) = cuberoot (3^2 a)
2) To make the powers divisuble by three (because it's a CUBE root), you need one more 3 (to make 3^3) and two more a's (to make a^3). So you would normally multiply by cuberoot(3a^2).

Multiplying cuberoot(9a) times cuberoot (3a^2) would give cuberoot(27a^3), which is "doable", and equals 3a.

Had you used this technique on your problem (without first simplfying the fraction), it would have flowed as follows:

cuberoot(3a^2)
-.-.-.-.-.-.-.-.-.-.-.
cuberoot(9a)




cuberoot(3a^2) * cuberoot(3a^2)
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.
cuberoot(9a)*cuberoot(3a^2)


cuberoot(9a^4)
-.-.-.-.-.-.-.-.-.-.-.
cuberoot(27a^3)


Now the denominator is "dobale".


cuberoot(9a^4)
-.-.-.-.-.-.-.-.-.-.-.
3a


However, this can be simplified further by simplifying the radical upstairs AND by reducing the fraction (this is the price you pay for NOT reducing the fraction first, as puzzling did).

cuberoot(a^3)*cuberoot(9a)
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
3a

a*cuberoot(9a)
-.-.-.-.-.-.-.-.-.-.-.
3a

cuberoot(9a)
-.-.-.-.-.-.-.-.-.
3

Answer 4

You can factor them out to be products of cube roots:

(³√3a * ³√a) / (³√3a * ³√3) = ³√a / ³√3 = ³√(a/3)

Answer 5

=cube root of (a/3)