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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.

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    $\begingroup$ 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 $$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha).$$ $\endgroup$ – Jyrki Lahtonen Dec 17 '17 at 10:49
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\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?

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  • $\begingroup$ Thanks. I have started to think what am I doing for last few days. Thanks.. :) $\endgroup$ – CodeBlooded Dec 17 '17 at 10:50
  • $\begingroup$ Yeah, but I have to wait for at least 4 minutes before I mark it correct. Don't worry I will do it soon. :) $\endgroup$ – CodeBlooded Dec 17 '17 at 10:52
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$$\alpha^5+\beta^5+\gamma^5=(\alpha+\beta+\gamma)^5-5(\alpha+\beta+\gamma)^3(\alpha\beta+\alpha\gamma+\beta\gamma)+5(\alpha\beta+\alpha\gamma+\beta\gamma)^2+$$ $$+5(\alpha+\beta+\gamma)^2\alpha\beta\gamma-5(\alpha\beta+\alpha\gamma+\beta\gamma)\alpha\beta\gamma=1+5\cdot(-1)=-4.$$

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

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

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

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