Sum of fifth power of the roots of equation $x^3-x^2+1=0$

The equation $x^3-x^2+1=0$ has three roots $\alpha$, $\beta$ and $\gamma$. Find the value of $\alpha^5 + \beta^5 + \gamma^5$

I tried it this way: $x^3=x^2-1$

$\alpha + \beta + \gamma = 1$

$\alpha \cdot \beta \cdot \gamma = -1$

$\alpha \cdot \beta + \beta \cdot \gamma + \alpha \cdot \gamma = 0$

So, $\alpha^3=\alpha^2-1$

$\alpha^5=\alpha^4-\alpha^2$

And similarly for $\beta$ and $\gamma$ Now I did add them but I am unable to find something useful in it.

• Keep going! You know that $\alpha^3=\alpha^2-1$. Therefore $$\alpha^4=\alpha(\alpha^2-1)=\alpha^3-\alpha=\alpha^2-\alpha-1.$$ Similarly you can write $\alpha^5$ in terms of even lower degree powers. Then use Vieta relations and $$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha).$$ – Jyrki Lahtonen Dec 17 '17 at 10:49

$$\alpha^5+\beta^5+\gamma^5=(\alpha+\beta+\gamma)^5-5(\alpha+\beta+\gamma)^3(\alpha\beta+\alpha\gamma+\beta\gamma)+5(\alpha\beta+\alpha\gamma+\beta\gamma)^2+$$ $$+5(\alpha+\beta+\gamma)^2\alpha\beta\gamma-5(\alpha\beta+\alpha\gamma+\beta\gamma)\alpha\beta\gamma=1+5\cdot(-1)=-4.$$
$$\alpha^2+\beta^2+\gamma^2=(\underbrace{\alpha+\beta+\gamma})^2-2(\underbrace{\alpha\beta+\beta\gamma+\gamma\alpha})=?$$
$$\alpha^4+\beta^4+\gamma^4=(\underbrace{\alpha^2+\beta^2+\gamma^2})^2-2(\underbrace{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2})$$
Now $$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\underbrace{\alpha\beta+\beta\gamma+\gamma\alpha})^2-2\underbrace{\alpha\beta\gamma}(\underbrace{\alpha+\beta+\gamma})$$