Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$ 
If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that
$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$

I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know how simplify it to 2.
I try:
Let $a=F_n$, $b=F_{n+1}$ and $c=F_{n+2}$
$a^4+b^4+c^4+2(ab)^2+2(ac)^2+2(bc)^2=2a^4+2b^4+2c^4$
$2(ab)^2+2(ac)^2+2(bc)^2=a^4+b^4+c^4$
I am not sure, what to do next. Can anyone help by completing the prove?

Setting $x=a$, $y=b$, and $z=-(a+b)$ we have to evaluate $\frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4}$
Since $x+y+z= 0$, $x,y,z$ are roots of $t^3 + t(\sum xy)-xyz = 0$
Hence $x,y,z$ satisfy $t^4 + t^2(\sum xy)-xyzt = 0$
Setting $t=x,y,z$ successively and adding the resulting equations we obtain
$\sum x^4 +\sum x^2 \sum xy -xyz \sum x = 0 \implies \sum x^4 +\sum x^2 \sum xy  = 0$ 
Since $\sum x = 0$ we get that $2\sum xy = - \sum x^2$
Now its easy to see that $\frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4} = 2$
