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Suppose I've got this:

In[13]:= Expand[(a + b) (b + c) (c + a)]

Out[13]= a^2 b + a b^2 + a^2 c + 2 a b c + b^2 c + a c^2 + b c^2

And I want to collect only terms involving a^2. In other words, I want the following output:

a^2(b + c) + a b^2 + 2 a b c + b^2 c + a c^2 + b c^2

How can I do this? If I use the following:

Collect[%, a^2]

Then it simply groups terms into the highest power of a, even if the highest term is less than 2. So it results in this:

In[14]:= Collect[%, a^2]

Out[14]= b^2 c + b c^2 + a^2 (b + c) + a (b^2 + 2 b c + c^2)

Ideally, I would like to extend this further to collect all a^2, b^2, and c^2 in one expression. So that running my command would transform the original fully expanded expression into the following:

a^2(b + c) + b^2(a+c) + c^2(a+b) + 2 a b c

In a single command. It this possible?

Zachary Turner
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    FYI (its not the answer to your question but...) another symmetric form would be given by SymmetricReduction[Expand[(a + b) (b + c) (c + a)] , {a, b, c}][[1]] – chris Nov 15 '12 at 09:32
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3 Answers3

4

Introducing dummy variables will do the job:

Collect[Expand[(a + b) (b + c) (c + a)]
  /. {a^2 -> x, b^2 -> y, c^2 -> z}, {x, y, z}]/. {x -> a^2, y -> b^2, z -> c^2}

2 a b c + (a + b) c^2 + b^2 (a + c) + a^2 (b + c)

If you will have more variables in the future, it probably makes sense to rewrite this:

P = {a, b, c}; Q = {x, y, z};
Collect[Expand[(a + b) (b + c) (c + a)] 
   /. MapThread[#1^2 -> #2 &, {P, Q}], Q] /. MapThread[#2 -> #1^2 &, {P, Q}]

2 a b c + (a + b) c^2 + b^2 (a + c) + a^2 (b + c)

Vitaliy Kaurov
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3

How about Coefficient?

expr = Expand[(a + b) (b + c) (c + a)]
term = Coefficient[expr, a^2]*a^2 (* get the coefficient and multiply it with the variable *)
rest = (expr - term) // Expand (* expand the rest *)
term + rest

a^2 b + a b^2 + a^2 c + 2 a b c + b^2 c + a c^2 + b c^2

a^2 (b + c)


a b^2 + 2 a b c + b^2 c + a c^2 + b c^2


a b^2 + 2 a b c + b^2 c + a c^2 + b c^2 + a^2 (b + c)

The final line contains the a^2(b + c) term at the end.

István Zachar
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0

Your first question is quite different from the second one. To address the first one you might do as follows:

pos = Position[Collect[expr, a], a^2*b_][[1, 1]]

this yields 3. Then 

Collect[expr, a][[pos]]

yielding a^2 (b + c)

Alexei Boulbitch
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