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When I try this

Simplify[(1-a)*(1/(1+x*m))^(1-a)(1+m)^(1-a) + 

the output is

$$\frac{(1-a) (1+m)^{1-a} \left(\left(\frac{1}{1+m x}\right)^{-a}+m x (m x)^a \left(\frac{m x}{1+m x}\right)^{-a}\right)}{1+m x}$$

But a little hand-done algebra shows this reduces to $(1-a) (1+m)^{1-a} (1+mx)^a$.

I tried FullSimplify but get the same results. Is there something I can do to get Mathematica find this simpler form?

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up vote 12 down vote accepted

Adding assumptions does that:

In[16]:= Simplify[(1 - a)*(1/(1 + x*m))^(1 - a) (1 + m)^(1 - a) + (1 -
      a)*(x*m)^a*((x*m)/(1 + x*m))^(1 - a)*(1 + m)^(1 - a), 
 1 + m x > 0 && a > 0]

Out[16]= -(-1 + a) (1 + m)^(1 - a) (1 + m x)^a

The issue is that (1/x)^a != x^(-a) for all complex $x$ and $a$. Indeed, let $x=-1$ and $a=\frac{1}{2}$:

In[17]:= With[{x = -1, a = 1/2}, {(1/x)^a, x^-a}]

Out[17]= {I, -I}
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Thanks. I'm new to Mathematica, but I suppose it makes sense that it answers the question I ask rather than the question I mean to ask. – itzy Nov 15 '12 at 21:21
It has become a habit for me to use FullSimplify[expr, _\[Element]Reals] whenever I simplify anything not explicitly complex. – ssch Nov 15 '12 at 21:22

Another approach is to use PowerExpand first which gets you to the same answer.

Simplify[PowerExpand[(1-a)*(1/(1+x*m))^(1-a)(1+m)^(1-a) + 
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