Question 395152
{{{(root(3, 9)*root(3, 6))/(root(6, 2)*root(6, 2))}}}
First we can use a property of radicals, {{{root(a, p)*root(a, q) = root(a, p*q)}}}, to multiply in both the numerator and the denominator:
{{{root(3, 9*6)/root(6, 2*2)}}}
or
{{{root(3, 54)/root(6, 2^2)}}}
I left the radicand in the denominator in the form of {{{2^2}}} for reasons that will become clear soon.
Next we use fractional exponents in the denominator to help us simplify further. Using the pattern {{{root(a, p^b) = a^(b/a)}}} we can rewrite the denominator as:
{{{2^(2/6)}}}
We can see that this fraction will reduce to:
{{{2^(1/3)}}}
Writing this back in radical form we get:
{{{root(3, 2^1)}}}
or
{{{root(3, 2)}}}
Now our fraction looks like:
{{{root(3, 54)/root(3, 2)}}}
Now we can use another property of radicals, {{{root(a, p)/root(a, q)  = root(a, p/q)}}} to combine the two radicals into one:
{{{root(3, 54/2)}}}
which simplifies to:
{{{root(3, 27)}}}
And since {{{3^3 = 27}}}, this cube root becomes a 3!