<s>[INST] If $\frac1x+\frac1y+\frac1z=0, xyz \neq 0$, $\sqrt[3]{\tiny\frac{x^9+y^9+z^9-3xyz(x^6+y^6+z^6)+6x^3y^3z^3}{x^6+y^6+z^6-3x^2y^2z^2}} =?$ If $\frac1x+\frac1y+\frac1z=0,$ and $xyz \neq 0$, What is $$K =\sqrt[3]{\frac{x^9+y^9+z^9-3xyz(x^6+y^6+z^6)+6x^3y^3z^3}{x^6+y^6+z^6-3x^2y^2z^2}} ?$$
Source: Lumbreras Editors
I found this way:
$  (xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$
$ \Rightarrow (xy)^2+(yz)^2+(zx)^2 =-2xyz(x+y+z) $
For Gauss:
$ (xy)^3+(yz)^3+(zx)^3=3x^2y^2z^2$
Whit:
$(x+y+z)(\underbrace{xy+yz+zx}_{0})=(x+y)(y+z)(z+x)+xyz$
$ ⇒(x+y)(y+z)(z+x)    =-xyz$
And
$ (x+y+z)^2 =x^2+y^2+z^2+2(\underbrace{xy+yz+zx}_{0})$
$ ⇒ (x+y+z)^2 =x^2+y^2+z^2$
$ ⇒ (x+y+z)^6 = x^6+y^6+z^6+3(x^2+y^2+z^2)\underbrace{x^2y^2+y^2z^2+z^2x^2}_{-2xyz(x+y+z)})\\                               
  -3x^2y^2z^2$
$ ⇒ (x+y+z)^3[(x+y+z)^3+6xyz]  = x^6+y^6+z^6-3x^2y^2z^2$
\intertext{Trinomio al cubo:}
$(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(\underbrace{xy+yz+zx}_{0})-3xyz $
$ ⇒ (x+y+z)^3=x^3+y^3+z^3-3xyz $
Then:
$ (x+y+z)^6=x^6+y^6+z^6+6x^2y^2z^2+3[x^3(y^6+z^6)+y^3(x^6+z^6)+z^3(x^6+y^6)]$
So far I got, I didn't know what else to do
 [/INST] Use the identity:
$$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$
Then the denominator, $A$, inside the square root is:
\begin{equation}A = (x^9+y^9+z^9-3x^3y^3z^3) -3xyz(x^6+y^6+z^6 - 3x^2y^2z^2)\end{equation}
The first summand is then further decomposed as:
\begin{align}
x^9+y^9+z^9-3x^3y^3z^3 = (x^3+y^3+z^3)(x^6+y^6+z^6 - x^3y^3-y^3z^3-z^3x^3)
\end{align}
However, the same identity gives us:
$$\sum\dfrac{1}{x^3} = \dfrac{3}{xyz}\iff \sum x^3y^3 = 3x^2y^2z^2.$$
Therefore, the cube of your fraction is:
$$\dfrac{\sum x^9 - 3x^3y^3z^3}{\sum x^6 - 3x^3y^2z^2} - 3xyz = x^3+y^3+z^3 - 3xyz.$$
Now, let $x+y+z=p$ and $xyz = r$ with $xy+yz+zx = 0.$ Then,
$$x^3+y^3+z^3 - 3xyz = (x+y+z)((x+y+z)^2-3xy-3yz-3zx) = p^3.$$
This means that your fraction is equal to:
$$p = x+y+z.$$
</s>