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I have studied that:

For the polynomial $ax^3+bx^2+cx+d=0$, with roots $\alpha, \beta, \gamma$:

We have:

$$\begin{align} & \alpha + \beta + \gamma = -\frac ba \\ & \alpha\beta + \beta\gamma + \alpha\gamma = \frac ca \\ & \alpha\beta\gamma = -\frac da \end{align} $$

A question asks to prove for: $x^3-px-q=0$ that:

$$\begin{align} & \alpha^2 + \beta^2 + \gamma^2 = 2p \\ & \alpha^3 + \beta^3 + \gamma^3 = 3q \end{align}$$

I looked around the Internet but couldn't find a way to prove this. How would I go around doing this?

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$$\left(\sum\alpha\right)^2=\sum\alpha^2+2\sum\alpha\beta$$

$$\iff\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta$$


$$a\alpha^3=-b\alpha^2-c\alpha-d$$

$$\implies a\sum\alpha^3=-b\sum\alpha^2-c\sum\alpha-3d$$

See also : Newton's Sums

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  • $\begingroup$ I was able to understand Newton's Sums from the link you provided, thank you. But what exactly do you mean by $\sum\alpha$? $\alpha$ is just one of the roots. $\endgroup$ – Youssef Moawad Sep 12 '14 at 11:40
  • $\begingroup$ @YoussefSami, Here $$\sum\alpha=\alpha+\beta+\gamma$$ $\endgroup$ – lab bhattacharjee Sep 12 '14 at 11:43
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The squared sum ($0$) minus twice the sum of products ($-p)$ is the sum of squares, $0-2(-p)=2p$.

From the equation, the cube equals $p$ times the value plus $q$. By summing, $p\cdot(0)+3q=3q$.

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Let $f(x)=x^3−px−q=0 $ and assume that $\alpha, \beta,\gamma$ are the roots of the polynomial $ f(x)$ so \begin{align} \alpha +\beta +\gamma =0\\ \alpha\beta +\beta\gamma +\gamma\alpha =-p\\ \alpha\beta\gamma=q \end{align} Consider, \begin{align} (\alpha +\beta +\gamma)^2 &= & 0\\ \implies \alpha^2+\beta^2+\gamma^2+2(\alpha\beta +\beta\gamma +\gamma\alpha) & =& 0\\ \implies \alpha^2+\beta^2+\gamma^2=-2(\alpha\beta +\beta\gamma +\gamma\alpha)=2p. \end{align} Similarly, we know that $$ \alpha +\beta +\gamma =0 \implies \alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma=3q. $$ (This can also be shown by the similar argument that we have done in the square. Just consider $(\alpha +\beta +\gamma)^3$ and expand it you will get the result that $\alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma $)

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