# Math Help - Solve the equation

1. ## Solve the equation

Solve the equation: $a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2+4(a+b+c)+12=6abc+ 4(ab+bc+ca)$

2. Originally Posted by james_bond
Solve the equation: $a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2+4(a+b+c)+12=6abc+ 4(ab+bc+ca)$
solve for what?

3. Originally Posted by skeeter
solve for what?
For $a$, $b$ and $c$. I've solved for the case $a=b=c$ and got two solutions. I'm pretty sure that there are no other solutions but I can't prove it.

4. So if $a=b=c$ the equation becomes:
$3a^4+3a^2+12a+12=6a^3+12a^2\Leftrightarrow a^4-2a^3-3a^2+4a+4=0\Leftrightarrow$ $(-2 + a)^2 (1 + a)^2=0$
So $a=b=c=2$ and $a=b=c=-1$ are solutions.

5. No ideas how to prove that there are no other solutions? Maybe this equation is an inequality and equality holds iff $a=b=c$. If so what's the inequality and how to prove it?

6. The equation can be written as

$(ab-c-2)^2+(ac-b-2)^2+(bc-a-2)^2=0$

Then

$\left\{\begin{array}{lll}ab-c=2\\ac-b=2\\bc-a=2\end{array}\right.$

Can you solve the system?

7. Thanks for your help but unfortunately I can't solve the equation system. I got $ab+b=2$ and similarly the other two but I can't get further. Could you help me out, please? Thank you!

8. Originally Posted by red_dog

$\left\{\begin{array}{lll}ab-c=2\\ac-b=2\\bc-a=2\end{array}\right.$
Substract the second equation from the first:

$(b-c)(a+1)=0$

1. If $b=c$, replace c with b in the first and the third equation:

$\left\{\begin{array}{ll}ab-b=2\\b^2-a=2\end{array}\right.$

From the second equation $a=b^2-2$

Replace a in the first equation: $b^3-3b-2=0\Leftrightarrow (b+1)^2(b-2)=0$

If $b=-1\Rightarrow a=-1, \ c=-1$

If $b=2\Rightarrow a=2, \ c=2$

2. If $a=-1$, replace a in the first and the third equation:

$\left\{\begin{array}{ll}b+c=-2\\bc=1\end{array}\right.\Rightarrow b=c=-1$