Suma de quinto poder de las raíces de la ecuación $x^3-x^2+1=0$

Question

El $x^3-x^2+1=0$ de la ecuación tiene tres raíces $\alpha$, $\beta$ y $\gamma$. Encuentra el valor de $\alpha^5 + \beta^5 + \gamma^5$

Lo probé de esta forma: $x^3=x^2-1$

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

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

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

Así, $\alpha^3=\alpha^2-1$

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

Y semejantemente para $\beta$ y $\gamma$ ahora añadirles pero soy incapaz de encontrar algo útil en él.

Answer 1

¿\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 {Alinee el} puede tomar desde aquí?

Answer 2

$$\alpha^5+\beta^5+\gamma^5=(\alpha+\beta+\gamma)^5-5(\alpha+\beta+\gamma)^3(\alpha\beta+\alpha\gamma+\beta\gamma)+5(\alpha\beta+\alpha\gamma+\beta\gamma)^2+$ $$$+5(\alpha+\beta+\gamma)^2\alpha\beta\gamma-5(\alpha\beta+\alpha\gamma+\beta\gamma)\alpha\beta\gamma=1+5\cdot(-1)=-4.$ $

Answer 3

$$\alpha^2+\beta^2+\gamma^2=(\underbrace{\alpha+\beta+\gamma})^2-2(\underbrace{\alpha\beta+\beta\gamma+\gamma\alpha})=?$$

$$\alpha^4+\beta^4+\gamma^4=(\underbrace{\alpha^2+\beta^2+\gamma^2})^2-2(\underbrace{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2})$$

Ahora $$\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})$ $