Question 837613
<pre>
{{{matrix(2,1,"", 3m^(2/3)-4m^(-1/3))}}} 

We can easily factor {{{matrix(2,1,"",m^(-1/3))}}} out of the second
term.  So the only hitch is in factoring {{{matrix(2,1,"",m^(-1/3))}}} out of the first term.  

Let's let P be the power of m that we leave inside the parentheses
when we factor {{{matrix(2,1,"",m^(-1/3))}}} out of the first term.
That is, suppose P is such that the original expression factors 
like this:

{{{matrix(2,1,"", m^(-1/3)(3m^P-4))}}}

Now let's multiply that back out by adding exponents of m 
on the first term:

{{{matrix(2,1,"", 3m^(-1/3+P)-4m^(-1/3))}}}

That has to equal the power of m in the first term of the original:

{{{matrix(2,1,"", 3m^(2/3)-4m^(-1/3))}}} 

So we must have {{{-1/3+P=2/3}}}.  Solve that for P
                {{{P=2/3+1/3}}}
                {{{P=3/3}}}
                {{{P=1}}}

So we replace P by 1 in

{{{matrix(2,1,"", m^(-1/3)(3m^P-4))}}}

and get:

{{{matrix(2,1,"", m^(-1/3)(3m^1-4))}}}

or just:

{{{matrix(2,1,"", m^(-1/3)(3m-4))}}}


Edwin</pre>