If a,b, and c is real, and following the equation, solve for 
If $a,b,c$ are real and aren't 0, that complete the following equations
$a^2 + a = b^2$
$b^2 + b = c^2$ 
$c^2 + c = a^2$
Solve for $(2a+c)(2b+a)(2c+b)$,

What I've done: 


*

*Adding all of the equation of it and substact with $(a^2 + b^2 + c^2)$ of both side of the equation
$a+b+c=0$ 

*Substituting $a = -(b+c); b=-(c+a); c=-(a+b)$ to equation $(1),(2),(3)$ respectively:
$c^2 -b -c  + 2bc= 0$
$a^2 - c -a + 2ac = 0$
$b^2 -b -a +2ab = 0$


*Expanding the equation into $4a^2c+2a^2b+9acb+4ab^2+2ac^2+4c^2b+2cb^2$
then re-arranged by it's constant
$4(a^2c+ab^2+c^2b)+2(a^2b+ac^2+cb^2)+9abc$
focused on
$2(a^2b+ac^2+cb^2)$
which can be written as 
$2((a)ab+(c)ac+(b)cb)$
and subsituting $a, b, c$ that inside parentheses
$2(-(b+c)(ab)-(a+b)ac-(a+b)cb)
-2(ab^2 + a^2c + cb^2 +3abc)$


removing $3abc$, and $Cb^2$ from the parentheses
$-2(ab^2 + a^2c) - 6abc -cb^2$
so I get
$(4-2)(a^2c+ab^2+cb^2)+9abc-6abc$
$2(a^2c+ab^2+cb^2)+3abc$
But that's seems I can go to, I didn't find anything that could be useful anymore

Adding all equations we get
$$a+b+c=0$$ and now we can eliminate the other variables, we obtain: $$3a^4-3a^2+a=0$$ and so on.
With $$c=-a-b$$ we get
$$b^2+b=(a+b)^2$$ and
$$(a+b)^2-(a+b)=a^2$$
From the equation $$b^2+b=a^2+b^2+2ab$$ we get
$$a^2+2ab-b=0$$ and $$b=\frac{a^2}{1-2a}$$ Can you proceed?
And we get $a$ from
$$a^2+a=\left(\frac{a^2}{1-2a}\right)^2$$
