+24 Hi I would appreciate it if someone helps me solve the question: Factor \((ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3\)as much as possible. Thanks!!
+1938 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 \(3abc(a+b)(b+c)(c+a)\)
Thanks! :)
