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           In Cap. 8 (Miscellaneous Problems) at page 236, Problem 47

 SOLUTION CiP

           The numbers $x,\;y,\;z$ are the roots of a third degree equation, with unknown $w$

(1)                     $w^{3}-(x+y+z)\cdot w^{2}+(xy+yz+zx)\cdot w -xyz=0$

 (according to Vieta's formulas). From the given conditions, and we can write the second one $\frac{yz+zx+xy}{xyz}=0$, the equation (1) is

(2)                       $w^{3}-x \cdot y \cdot z=0$.

From (2) we get

$w^{4}=xyz \cdot w$,          $w^{5}=xyz \cdot w^{2}$          $w^{6}=xyz \cdot w^{3}$.

 

We put $x,\; y,\; z$ in turn instead of $w$ in the last relation, we sum up the obtained results in and we get

$\sum x^{6}=xyz \cdot \sum x^{3}$.

 It follows from here $\frac{x^{6}+y^{6}+z^{6}}{x^{3}+y^{3}+z^{3}}=xyz$.

$\blacksquare$ 

 

      REMARK CiP

           Numbers that check the given conditions must be complex numbers. (Because $\sum x^{2}=(\sum x)^{2}-2 \cdot \sum xy =0$ and $xyz \neq 0$.)

           From relationships

$$\begin{cases}x+y=-z\\\frac{1}{x}+\frac{1}{y}=-\frac{1}{z} \end{cases}$$

 we deduce  

$x+y=-z$,          $x \cdot y =z^{2}$

so $x$ and $y$ are the  roots of the quadratic equation with unknown $t$

$t^{2}+z \cdot t +z^{2}=0$

 that is $\frac{-1\pm\imath \sqrt{3}}{2}\cdot z$. So the general family of solutions of the given equations is

$(\frac{-1+\imath \sqrt{3}}{2}\cdot z\;,\; \frac{-1-\imath \sqrt{3}}{2} \cdot z\;,\;z)\;\;\;z \in \mathbb{C}\setminus \{ 0 \}$.

 $\square$