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.
 A: \begin{align}\alpha^5+\beta^5+\gamma^5&=\alpha^4-\alpha^2+\beta^4-\beta^2+\gamma^4-\gamma^2\\&=\alpha^3-\alpha-\alpha^2+\beta^3-\beta-\beta^2+\gamma^3-\gamma-\gamma^2\\&=\alpha^2+1-\alpha-\alpha^2+\beta^2+1-\beta-\beta^2+\gamma^2+1-\gamma-\gamma^2\\&=3-(\alpha+\beta+\gamma).\end{align}Can you take it from here?
A: $$\alpha^5+\beta^5+\gamma^5=(\alpha+\beta+\gamma)^5-5(\alpha+\beta+\gamma)^3(\alpha\beta+\alpha\gamma+\beta\gamma)+5(\alpha\beta+\alpha\gamma+\beta\gamma)^2+$$
$$+5(\alpha+\beta+\gamma)^2\alpha\beta\gamma-5(\alpha\beta+\alpha\gamma+\beta\gamma)\alpha\beta\gamma=1+5\cdot(-1)=-4.$$
A: $$\alpha^2+\beta^2+\gamma^2=(\underbrace{\alpha+\beta+\gamma})^2-2(\underbrace{\alpha\beta+\beta\gamma+\gamma\alpha})=?$$
$$\alpha^4+\beta^4+\gamma^4=(\underbrace{\alpha^2+\beta^2+\gamma^2})^2-2(\underbrace{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2})$$
Now $$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\underbrace{\alpha\beta+\beta\gamma+\gamma\alpha})^2-2\underbrace{\alpha\beta\gamma}(\underbrace{\alpha+\beta+\gamma})$$
