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avatar+1897 
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I know this is a late response, but I just realized the answer to this question. 

Now, even though the blue checkmark has been given, signifying the right answer, I don't believe the anwer was correct. 

 

Let me explain my reasoning. 

First off, the final answer achieved in the previous response (no knock, you almost had the right answer) was

\(\left(a\:+\:b\:+\:c\right)\left(ab\:+\:bc\:+\:ca\right)\) This simplifies to \(a^2b+a^2c+ab^2+b^2c+bc^2+ac^2+3abc\)

 

This can't be right because note that in the original problem, there are powers of 3 for variables a, b, c, while in this, there are none. 

 

Now, let's me explain my tactic. Hang with me, it's quite tedious. 

Now first off, let's set a multivariable function to deal with this question, 

First, let's let \(f(a,b,c) = (ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3\). (This will come into handy later)

 

This following step is not necessary, as you can just easily follow through, but expanding everything, which is quite tedious, we get

\(3a^3b^2c+3a^2b^3c+3a^3bc^2+3a^2bc^3+3ab^3c^2+3ab^2c^3+6a^2b^2c^2\)

 

Now, let's note that every term is divisble by 3abc. Factoring this out, we get

\(f(a, b, c,)=3abc(b^2c+a^2c+a^2b+ab^2+2abc+bc^2+ac^2)\)

 

The following steps are complicated, but I do believe this was explained in an AOPS course, which is where this problem came from.

(I know because I had the same problem while taking the course myself, it was a writing problem)

 

Let's note the following ideas. 

If we let\( c=-a\), we find that \(f(a,b,-a) =0\), meaning that a+c is a factor of \(f(a,b,c)\)

If we let \(-a = b\), we also find that \(f(a, -a, c) = 0\). This also means that a+b is a factor. 

If we let b = -c,  \(f(a, -c, c) = 0\). This also means that b+c is a factor. 

 

This also means that \((a+b)(b+c)(c+a)\)should be in the final factorization. 

Wait a sec! Notice something really quickly!

\((a+b)(b+c)(c+a) = b^2c+a^2c+a^2b+ab^2+2abc+bc^2+ac^2\), which means that we can replace what we had in parethensis in the original facorization with \((a+b)(b+c)(c+a)\)

 

Thus, our final factorization is \(3abc(a+b)(b+c)(c+a)\)

 

Thus, our final answer is \(3abc(a+b)(b+c)(c+a)\)

 

Sorry for the late response. This should be correct, but correct me if it is not. 

 

Thanks! :)

Jul 22, 2024

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