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

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- Mar 7th 2009, 10:32 PMjames_bondSolve the equation
Solve the equation: $\displaystyle a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2+4(a+b+c)+12=6abc+ 4(ab+bc+ca)$

- Mar 8th 2009, 05:17 AMskeeter
- Mar 8th 2009, 06:25 AMjames_bond
- Mar 8th 2009, 06:55 AMjames_bond
So if $\displaystyle a=b=c$ the equation becomes:

$\displaystyle 3a^4+3a^2+12a+12=6a^3+12a^2\Leftrightarrow a^4-2a^3-3a^2+4a+4=0\Leftrightarrow$ $\displaystyle (-2 + a)^2 (1 + a)^2=0$

So $\displaystyle a=b=c=2$ and $\displaystyle a=b=c=-1$ are solutions. - Mar 9th 2009, 10:00 AMjames_bond
No ideas how to prove that there are no other solutions? Maybe this equation is an inequality and equality holds iff $\displaystyle a=b=c$. If so what's the inequality and how to prove it?

- Mar 10th 2009, 06:56 AMred_dog
The equation can be written as

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

Then

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

Can you solve the system? - Mar 12th 2009, 06:52 AMjames_bond
Thanks for your help but unfortunately I can't solve the equation system. I got $\displaystyle ab+b=2$ and similarly the other two but I can't get further. Could you help me out, please? Thank you!

- Mar 12th 2009, 12:57 PMred_dog
Substract the second equation from the first:

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

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

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

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

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

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

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

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

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