<s>[INST] Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.
 [/INST] Note: Using Newton's identities, we can calculate the below expressions more easily, following the easy recursive definition.
But, your idea of writing as roots of third degree polynomial works I believe, but requires some work and we show that here:
Let $\displaystyle x,y,z$ be roots of $\displaystyle t^3 + at - b = 0$. We have that $\displaystyle a = xy+yz+zx$ and $\displaystyle b = xyz$.
Since $\displaystyle t^3 = b - at$, multiply by $\displaystyle t^4$ we get $\displaystyle t^7= bt^4 - a t^5$.
Setting $\displaystyle t=x,y,z$ in turn and adding gives us

$\displaystyle x^7 + y^7 + z^7 = b(x^4 + y^4 + z^4) - a(x^5 + y^5 + z^5)$

Similar to above, we get $\displaystyle t^5 = bt^2 - a t^3$, setting $\displaystyle t=x,y,z$ and adding gives us

$\displaystyle x^5 + y^5 + z^5 = b(x^2 + y^2 + z^2) - a(x^2 + y^3 + z^3)$.

Similarly we get

$\displaystyle x^3 + y^3 + z^3 = 3b$

We also have $\displaystyle (x+y+z)^2 = 0$, giving us

$\displaystyle x^2 + y^2 + z^2 = -2a$.

Thus

$\displaystyle x^5 + y^5 + z^5 = -2ab - 3ab = -5ab$.

Now $\displaystyle t^4 = bt - at^2$ and in a similar fashion we get

$\displaystyle x^4 + y^4 + z^4 = -a(x^2+y^2 +z^2) = 2a^2$
Thus $\displaystyle xyz(x^4 + y^4 + z^4) = 2a^2 b$
and $\displaystyle x^7 + y^7 + z^7 = 2a^2b - (-5a^2b) = 7a^2 b$.
Thus the given expression is $\displaystyle \frac{7}{2}$.

This approach can be used to generate identities.
For instance, show that

$\displaystyle 10(x^7 + y^7 + z^7) = 7(x^2 + y^2 + z^2)(x^5 + y^5 + z^5)$

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