Question 611997
<pre>
Was the problem given as a quotient of two {{{1/3}}} powers,
like this:?

{{{(3(81a^5b^10)^(1/3))/((3b^4)^(1/3)))}}} 

or was it given equivalently as a quotient of two cube roots, 
like this:?

{{{(3(root(3,81a^5b^10)))/(root(3,3b^4))}}}

Regardless, consider it as in the above form.
Separate the 3 coefficient of the numerator out in front as a 
coefficient of the entire fraction:

{{{3*(expr(root(3,81a^5b^10))/(root(3,3b^4)))}}}

A quotient of cube roots can be written as the cube root of
a quotient:

{{{3*(expr(root(3,81a^5b^10/(3b^4))))}}}

Under the cube root, we divide the 3 on the bottom into the
81 in the top, getting 27, and we divide the b<sup>4</sup> in the
bottom into the b<sup>10</sup> on top by subtracting exponents
getting b<sup>6</sup>.  and we have this:

{{{3*(expr(root(3,27a^5b^6)))}}}

We know the cube root of 27 is 3 so we can take the 27 out of the
cube root radical and put a 3 in front of the cube root radical.
We can also take the cube root of b<sup>6</sup> by dividing the exponent
6 by the index of the radical 3, getting b<sup>2</sup> in front of
the radical.  So we have

{{{3*(expr(3b^2*root(3,a^5)))}}}
  
We can multiple the 3's in front getting 9 and eliminate the
parentheses:

{{{9b^2*root(3,a^5)}}}

We haven't finished because we have an exponent 5 under a radical
that is larger than the index of the radical, 3, since it is a cube
root.  So we write the exponent 5 in terms of the largest possible
multiple of the radical index, 3.  The largest multiple of 3 that
does not exceed 5 is 3 itself.  Therefore we write the 5 exponent
as 3+2:

{{{9b^2*root(3,a^(3+2))}}}

Then we remember the rule for adding exponents when we multiply
in reverse and change a<sup>3+2</sup> to ała˛:

{{{9b^2*root(3,a^3a^2))}}}

Finally we can take the cube root of ał out on front of the radical
as just "a" and our final answer is:

{{{9ab^2*root(3,a^2))}}}

Edwin</pre>