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 ? 

We can use the following identity to solve the problem in a more efficient manner than remembering a gigantic identity:
$$a^3+b^3+c^3= \sigma_1^3-3\sigma_1\sigma_2+3\sigma_3$$
where $\sigma_1=a+b+c, \ \sigma_2=ab+bc+ca$, and $\sigma_3=abc$ are the Elementary Symmetric Polynomials. Next we find the cubed versions of the symmetric polynomials as follows
$$a^3b^3+b^3c^3+c^3a^3=(ab+bc+ca)^3-3(ab+bc+ca)(a^2bc+b^2ac+c^2ab)+3a^2b^2c^2$$
$$a^3b^3+b^3c^3+c^3a^3=\sigma_2^3-3\sigma_1\sigma_3+3\sigma_3^2$$
and $$a^3b^3c^3=\sigma_3^3$$
Now we can plug these cubed versions back into our original identity to get
$$a^9+b^9+c^9=(\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3)^3-3(\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3)(\sigma_2^3 - 3\sigma_1\sigma_3+3\sigma_3^2)+\sigma_3^9=3$$
You can then expand if you'd like to obtain your version, but I think I'll leave it here.
