If $\frac1x+\frac1y+\frac1z=0, xyz \neq 0$, $\sqrt[3]{\tiny\frac{x^9+y^9+z^9-3xyz(x^6+y^6+z^6)+6x^3y^3z^3}{x^6+y^6+z^6-3x^2y^2z^2}} =?$

Question

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

Answer 1

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: $$p = x+y+z.$$