If a,b, and c is real, and following the equation, solve for

Question

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:

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

2. Substituting $a = -(b+c); b=-(c+a); c=-(a+b)$ to equation $(1),(2),(3)$ respectively:

$c^2 -b -c + 2bc= 0$

$a^2 - c -a + 2ac = 0$

$b^2 -b -a +2ab = 0$

3. 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

Answer 1

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 $$a^2+a=\left(\frac{a^2}{1-2a}\right)^2$$

Answer 2

Hah!

$a + b + c = 0$

And $a = b^2 - a^2 = (b-a)(b+a) = (b-a)(-c) = c(a-b)$

$b = c^2 - b^2 = (c-b)(c+b) = (c-b)(-c) = a(b-c)$

$c = a^2 - c^2 = (a-c)(a+c) = (a-c)(-b) = b(c-a)$.

and $(2a + c) = (a +(a+c)) = (a-b) = \frac ac$

$(2b + a) = (b + (b + a)) = (b-c) = \frac ba$

$(2c + b) = (c + (c+b)) = (c -a) = \frac cb$

So

$(2a+c)(2b+a)(2c+b) = \frac ac\frac ba \frac cb = 1$.