May I know is there quick way to factorize multi-variables equations? I want to factorize such equation:
$$b^2c + bc^2 +a^2c+ac^2+a^2b+ab^2 +2abc$$ into product of linear factors.
May I know is there any quick way/trick to do so?I am very confused with such equation.  
Hello users, my main point here is what to think when we looking at factoring such such multi-variable questions. When I was doing this, I can't even start writing, since I really don't know what to do. Is there any formulae to remember? Or is just experience. I knew it might be very easy for some of you, but it is very difficult for me to factorize such equation. 
 A: I would write it like $$(ma+nb+kc)(m'a+n'b+k'c)(m''a+n''b+k''c)$$
Now, if we look it as a polynimial in $a$ we see that there is no $a^3$ so $mm'm''=0$ and we can assume $m''=0$. Similary we get $nn'n''=0$. Obviously $n''\ne 0$ since we can not take out $a$ in a starting expression. So we can assume $n'=0$ and similary $k=0$. So we have now $$(ma+nb)(m'a+k'c)(n''b+k''c)= \color{red}{mm'n''}a^2b+\color{red}{mm'k''}a^2c + ...$$ 
so $mm'n'' = mm'k''=1$ so $n'' = k''$ and similary $m'=k'$ and $m=n$. We have also  $mk'n''+m'nk''=2$. Clearly if all are $1$ it works so the answer is $$(a+b)(a+c)(b+c)$$ which is easy to verify.
A: There is a nice article on this, Gary Brookfield (2016) Factoring Forms, The American Mathematical Monthly,
volume 123 number 4, pages 347-362. The main theorem was proved by Aronhold in 1849. 
The general theorem, easy to state, is that a homogeneous cubic factors completely over the complexes if and only if the determinant of its Hessian matrix is a constant multiple of the original form. Here it is convenient to take exactly half the Hessian, 
$$
\frac{H}{2} =
\left(
\begin{array}{ccc}
b+c & a+b+c & a+b+c \\
a+b+c & a+c & a+b+c \\
a+b+c & a+b+c & a+b \\
\end{array}
\right)
$$
and the determinant of this is precisely the original polynomial. So, it factors completely, as the others have already shown. In this case, everything can be done with real numbers
In this case, the determinant calculation gives us extra information, as it says (take the original plynomial as $f$)
$$  f  = (b+c)(a+c)(a+b) + 2 (a+b+c)^3 - 2(a+b+c)(a+b+c)^2 =  (b+c)(a+c)(a+b) $$
I learned about this while attempting a different factoring problem
$$ (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right) $$
The roots of $\eta^3 - 3 \eta - 1 = 0$ are 
$$  A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B =   2 \cos  \left( \frac{5 \pi}{9} \right)  \approx -0.347 \; \; \; , C = 2 \cos      \left( \frac{ \pi}{9} \right)  \approx 1.879 \; \; \; . $$
We get identity
$$ \color{red}{(x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)} = \color{magenta}{(Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz)  }  $$ 
That one was at How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$
Here is one I made up yesterday for a different problem,
$$ f(x,y) = x^3-3x^2 y - 3 x y^2 + y^3 - 6x^2z -6 y^2z + 16 z^3 $$
In factoring, all coefficients are real, mostly rational but we do need to throw in a $\sqrt 3$ as part of some of the coefficients.
A: You can do it also like this $$a^2(b+c)+a(b^2+2bc+c^2)+ bc^2+c^2b=$$
$$a^2(b+c)+a(b+c)^2+bc(b+c)=$$
$$(b+c)(a^2+a(b+c)+bc)=$$
$$(b+c)(a^2+ab+ac+bc)=$$
$$(b+c)\Big(a(a+b)+c(a+b)\Big)=$$
$$(b+c)(a+b)(a+c)$$
A: If $a=-b$ we'll get $0$...
Thus, it's just $$(a+b)(a+c)(b+c).$$
Try to get it.
