Plz help asap!

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! :)