Mini-Challenge on a condition Hello I would like to purpose to you an enigma this is the following :

Let $a,b,c$ be positive real number find the condition on $abc$ and $ab+bc+ca$ and $a+b+c$ to have $a^9+b^9+c^9=3$ 

I have a solution using the following identity :
$$a^9+b^9+c^9= 3a^3b^3c^3−45abc(ab+bc+ca)(a+b+c)^4+ 54abc(ab+bc+ca)^2(a+b+c)^2−27a^2b^2c^2(ab+bc+ca)(a+b+c) + (a+b+c)^9−9(ab+bc+ca)(a+b+c)^7+ 9(ab+bc+ca)^4(a+b+c)−30(ab+bc+ca)^3(a+b+c)^3+ 18a^2b^2c^2(a+b+c)^3+ 27(ab+bc+ca)^2(a+b+c)^5+ 9abc(a+b+c)^6−9abc(ab+bc+ca)^3$$
But I would like a solution without this because it's too ugly .
So could you help me ? 

If you set $e_1=a+b+c,$ $e_2=ab+ac+bc,$ and $e_3=abc$ and you define a recurrence relation as follows:
\begin{eqnarray*}
x_0 &=& 3 \\
x_1 &=& e_1 \\
x_2 &=& e_1^2-2e_2 \\
x_{n+3} &=& e_1x_{n+2} - e_2x_{n+1} + e_3x_n
\end{eqnarray*}
then it can be shown that $x_n = a^n+b^n+c^n.$ In order to have $a^9+b^9+c^9=3,$ there must be $x_0,\ldots,x_9$ such that
$$\begin{pmatrix}
1 &&&&&&&&& \\
& 1 &&&&&&&& \\
&& 1 &&&&&&& \\
 -e_3 & e_2 & -e_1 & 1 &&&&&& \\
& -e_3 & e_2 & -e_1 & 1 &&&&& \\
&& -e_3 & e_2 & -e_1 & 1 &&&& \\
&&& -e_3 & e_2 & -e_1 & 1 &&& \\
&&&& -e_3 & e_2 & -e_1 & 1 && \\
&&&&& -e_3 & e_2 & -e_1 & 1 & \\
&&&&&& -e_3 & e_2 & -e_1 & 1  \\
&&&&&&&&& 1\end{pmatrix}
\begin{pmatrix}
x_0 \\x_1 \\x_2 \\x_3 \\x_4 \\x_5 \\x_6 \\x_7 \\x_8 \\x_9
\end{pmatrix}
=
\begin{pmatrix}
3 \\ e_1 \\ e_1^2-2e_2 \\0\\0\\0\\0\\0\\0\\0\\3
\end{pmatrix}
$$
In other words
$$\det
\begin{pmatrix}
1 &&&&&&&&&& 3\\
& 1 &&&&&&&&& e_1\\
&& 1 &&&&&&&& e_1^2-2e_2\\
 -e_3 & e_2 & -e_1 & 1 &&&&&&& \\
& -e_3 & e_2 & -e_1 & 1 &&&&&& \\
&& -e_3 & e_2 & -e_1 & 1 &&&&& \\
&&& -e_3 & e_2 & -e_1 & 1 &&&& \\
&&&& -e_3 & e_2 & -e_1 & 1 &&& \\
&&&&& -e_3 & e_2 & -e_1 & 1 && \\
&&&&&& -e_3 & e_2 & -e_1 & 1 & \\
&&&&&&&&& 1 & 3 \end{pmatrix} = 0
$$
