Solving equations

It can be verified that $x,y$ and $z$ are the roots of the equation $3u^3 - 9u^2 + 6u +2=0.$

It is equivalent to say that $u^3 = \frac{1}{3}\left(9u^2-6u-2 \right)$ . Multiply both sides of equation by $u$ , we have $u^4 = \frac{1}{3}\left(9u^3-6u^2-2u \right)$ .

Since $x,y$ and $z$ are the roots of the equation, it means that

$\large \begin{cases} x^4 = \frac{1}{3}\left(9x^3-6x^2-2x \right)\\ y^4 = \frac{1}{3}\left(9y^3-6y^2-2y \right)\\ z^4 = \frac{1}{3}\left(9z^3-6z^2-2z \right) \end{cases}$

Get the sum, we have $x^4+y^4+z^4 = \frac{1}{3}\left(9(7)-6(5)-2(3) \right) = 9$ .

Using the similar argument, we have $u^5 = \frac{1}{3}\left(9u^4-6u^3-2u^2 \right)$ and $x^5+y^5+z^5 = \frac{1}{3}\left(9(9)-6(7)-2(5) \right) = 9\frac{2}{3} \neq 11$ .