Question 173511
I'm not sure how I would slove anything, and an expression like this cannot be solved, only evaluated and simplified.


If you want to evaluate {{{sqrt(28y^3)}}} you could first factor the radicand, namely {{{28y^3}}} into its prime factors:


{{{28=2*2*7}}} and {{{y^3=y*y*y}}}


So the prime factorization of {{{28y^3}}} is {{{2*2*7*y*y*y}}}.


Now look for pairs of factors.


You have one pair of {{{2}}}s, and a pair of {{{y}}}s.


For each pair of factors, you can remove the pair from under the radical (the square root sign) and leave one of the pair of factors outside the radical.


Take out the pair of 2s, leaving {{{7*y*y*y}}} under the radical, and put one 2 outside the radical, thus:  {{{2sqrt(7*y*y*y)}}}.  Now do the same thing with the pair of {{{y}}}s:  {{{2y*sqrt(7*y)}}}.


That's the best that can be done with this one.