Answer:
Third choice from the top is the one you want
Step-by-step explanation:
This whole concept relies on the fact that if the index of a radical exactly matches the power under the radical, both the radical and the power cancel each other out. For example:
[tex]\sqrt[6]{x^6} =x[/tex] and another example:
[tex]\sqrt[12]{2^{12}}=2[/tex]
Let's take this step by step. First we will rewrite both the numerator and the denominator in rational exponential equivalencies:
[tex]\frac{\sqrt[4]{6} }{\sqrt[3]{2} }=\frac{6^{\frac{1}{4} }}{2^{\frac{1}{3} }}[/tex]
In order to do anything with this, we need to make the index (ie. the denominators of each of those rational exponents) the same number. The LCM of 3 and 4 is 12. So we rewrite as
[tex]\frac{6^{\frac{3}{12} }}{2^{\frac{4}{12} }}[/tex]
Now we will put it back into radical form so we can rationalize the denominator:
[tex]\frac{\sqrt[12]{6^3} }{\sqrt[12]{2^4} }[/tex]
In order to rationalize the denominator, we need the power on the 2 to be a 12. Right now it's a 4, so we are "missing" 8. The rule for multiplying like bases is that you add the exponents. Therefore,
[tex]2^4*2^8=2^{12}[/tex]
We will rationalize by multiplying in a unit multiplier equal to 1 in the form of
[tex]\frac{\sqrt[12]{2^8} }{\sqrt[12]{2^8} }[/tex]
That looks like this:
[tex]\frac{\sqrt[12]{6^3} }{\sqrt[12]{2^4} }*\frac{\sqrt[12]{2^8} }{\sqrt[12]{2^8} }[/tex]
This simplifies down to
[tex]\frac{\sqrt[12]{216*256} }{\sqrt[12]{2^{12}} }[/tex]
Since the index and the power on the 2 are both 12, they cancel each other out leaving us with just a 2! Doing the multiplication of those 2 numbers in the numerator gives us, as a final answer:
[tex]\frac{\sqrt[12]{55296} }{2}[/tex]
Phew!!!
