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 ? 

To make it less ugly, calculate the lower amounts instead, or setting new variables, for example:
${\begin{cases}a+b+c=x\\ab+bc+ca=y\\abc=z\\S_{n}=a^n+b^n+c^n\end{cases}}$
In fact, I have encountered this problem in a test that gives $x=3;y=-10;z=11$ and it asks me to find $S_{5}$. Here's how I attemped it (I have generalized it for $x;y;z$):
Step one:

$(a+b+c)(ab+bc+ca)=xy$
$\Rightarrow a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=xy$
$\Rightarrow a^2(b+c)+b^2(c+a)+c^2(a+b)=xy-3z$

Step two:

$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=x^2-2y$

Step three:

$(a+b+c)(a^2+b^2+c^2)=x(x^2-2y)$
$\Rightarrow a^3+b^3+c^3+a^2(b+c)+b^2(c+a)+c^2(a+b)=x(x^2-2y)$
$\Rightarrow a^3+b^3+c^3+xy-3z=x^3-2xy$
$\Rightarrow a^3+b^3+c^3=x^3-3xy+3z$

Step four:

$a^4+b^4+c^4$
$=(a^2+b^2+c^2)^2-2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$
$=(a^2+b^2+c^2)^2-2((ab+bc+ca)^2-2a^{2}bc-2ab^{2}c-2abc^{2})$
$=(a^2+b^2+c^2)^2-2((ab+bc+ca)^2-2abc(a+b+c))$
$=(x^2-2y)^2-2(y^2-2zx)$
$=x^4-4x^{2}y+2y^2+4zx$

Step five:

$(a^3+b^3+c^3)(ab+bc+ca)=y(x^3-3xy+3z)$
$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+abc(a^2+b^2+c^2)=x^3y-3xy^2+3yz$
$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+z(x^2-2y)=x^3y-3xy^2+3yz$
$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+x^2z-2yz=x^3y-3xy^2+3yz$
$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)=x^3y-x^2z-3xy^2+5yz$

Step six:

$(a+b+c)(a^4+b^4+c^4)=x(x^4-4x^{2}y+2y^2+4zx)$
$\Rightarrow a^5+b^5+c^5+a^4(b+c)+b^4(c+a)+c^4(a+b)=x^5-4x^3y+2xy^2+4x^2z$
$\Rightarrow a^5+b^5+c^5+x^3y-x^2z-3xy^2+5yz=x^5-4x^3y+2xy^2+4x^2z$
$\Rightarrow S_{5}=a^5+b^5+c^5=x^5-5x^3y+5xy^2+5x^2z-5yz$

You can continue similarly to $S_{9}$ to write the expression in terms of $x,y,z$, this will make the result much less ugly and also much easier to check. Thanks for taking time reading this long post.
