Question 253231
We start with fourth root (49*u^14).
We can rewrite this in fractional exponent form as
(i) {{{(49u^14)^(1/4)}}}
Now, we are looking for groups of 4 in order to pull out one set. So I will expand (i) to get
(ii) {{{7*7*u*u*u*u*u*u*u*u*u*u*u*u*u*u}}}
Now I look for groups of four. I get
(iii) {{{7*7*(u*u*u*u)*(u*u*u*u)*(u*u*u*u)*u*u}}}
I see 3 groups of 4 u's, so I can take out the 3 groups as u^3. Nothing else comes out, so we are left with the answer as:
--
{{{u^3*(7*7*u*u)^(1/4)}}}
or
{{{u^3*(49u^2)^(1/4)}}}