Question 418900
{{{root(4, 2/3)}}}
Part of simplifying radicals is eliminating:<ul><li>any radicals in denominators, and</li><li>any fractions within a radical.</li></ul> You have a fraction in a radical to eliminate.<br>
To eliminate a fraction within a radical, I like to multiply the numerator and denominator of that fraction by any number that will turn the denominator into a power that matches the type of root. With this expression, with its 4th root, we are looking to make the denominator a perfect 4th power. It is to your advantage to use the smallest number that will make the denominator a perfect 4th power. In this case, since the denominator is 3, a prime number, the lowest number that will turn a 3 into a perfect 4th power will be {{{3^3}}} which is 27:
{{{root(4, (2/3)(27/27))}}}
which simplifies to
{{{root(4, 54/81)}}}
Next we use a property of radicals, {{{root(a, p/q) = root(a, p)/root(a, q)}}}, separate the the numerator and denominator into their own 4th roots:
{{{root(4, 54)/root(4, 81)}}}
The denominator, since we know 81 is the product of 4 3's, simplifies:
{{{root(4, 54)/3}}}
There are no 4th power factors in 54 so we are finished simplifying.<br>
Note: If the expression had been {{{root(4, 3/4)}}} then we could but would not want to multiply the numerator nd denominator by {{{4^3}}}. This is so because 4 is not prime. 4 = 2*2. So all we need to get a perfect 4th power is two more 2's: 2*2 or 4. So
{{{root(4, 3/4) = root(4, (3/4)(4/4)) = root(4, 12/16) = root(4, 12)/root(4, 16) = root(4, 12)/2}}}
If we had used {{{4^3}}} we would have ended up with
{{{root(4, 192)/4}}}
This is correct, too. But it is not fully simplified. The numerator will simplify. And after that the fraction will reduce and (surprise, surprise) we end up with {{{root(4, 12)/2}}}! So you can see that when you don't turn the denominator into the lowest 4th power possible you end up with significant extra work.