Question 83550
<pre>

Stanbon's solution to your problem is incorrect.

Simplify and ractionalize all denominators:

{{{3b*sqrt(27a^5b) + 2a*sqrt(3a^3b^3)}}} 

There are no denominators to rationalize, so we just simplify this one

Break up what's under the radicals completely into prime factors:

Break up {{{27a^5b}}} into primes as {{{3*3*3*a*a*a*a*a*b}}}

Break up {{{3a^3b^3}}} into primes {{{3*a*a*a*b*b*b}}}

Substitute these under the radicals:

{{{3b*sqrt(3*3*3*a*a*a*a*a*b)}}} + {{{2a*sqrt(3*a*a*a*b*b*b)}}}

Since the index of the root is 2 (square root) we group all
like factors by twos (pairs) that we can like this.  We will
often have factors left over that would not pair up:

{{{3b*sqrt( (3*3)*3*(a*a)*(a*a)*a*b )}}} + {{{2a*sqrt( 3*(a*a)*a*(b*b)*b ) }}}

Each pair of like factors can be written as a square

{{{3b(sqrt(  (3^2)*3*(a^2)*(a^2)*a*b  )  )}}} + {{{2a(sqrt(3*(a^2)*a*(b^2)*b))  }}}

Take out each square from under the radical out front of the radical,
like this, leaving what didn't pair up under the radical:

{{{3b*3*a*a*sqrt(3*a*b)}}} + {{{ 2a*a*b( sqrt(3*a*b) )  }}}

Simplify what's in front of the radicals:

{{{9a^2b*sqrt(3ab)}}} + {{{2a^2b*sqrt(3ab)}}}

Now you can factor out GCF = {{{a^2b*sqrt(3ab)}}}

{{{a^2b*sqrt(3ab)}}}({{{9}}} + {{{2}}})

{{{a^2b*sqrt(3ab)11}}}

{{{ 11a^2b*sqrt(3ab) }}}

Edwin</pre>