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While reading about quadratic equations, I came across Newton's Identity formula which said we can express $\alpha^n+\beta^n$ in simpler forms but not given any explanation. They wrote $S_n=\alpha^n+\beta^n$ and plugged in the quadratic equation $f(x)=ax^2+bx+c$ to write: $$f(x)=aS_{n+1}+bS_n+cS_{n-1}$$ Also what if we want to find for any polynomial the value: $$\alpha^n-\beta^n$$ I went through articles on internet but they have some very complicated proofs. I want to understand how do we derive this in a simple manner and what have they exactly done with the quadratic equation? Some examples would make it very clear to understand. Thanks

UNAN
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    Please tell us what have you tried and where are you stuck –  Jul 10 '21 at 11:04
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    @JitendraSingh, I have read it for the first time and I haven't understood what the TB is trying to convey and want to know exactly how do we use this? Some examples would make it clear – UNAN Jul 10 '21 at 11:06
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    What is $f(x)$ first? Do we need Newton polynomials for the question, or this is just some misleading discussion? If we really need Newton polynomials, where is one such polynomial used in the question, best with clear notation?! And which is exactly the question? You want examples for what exactly? – dan_fulea Jul 10 '21 at 11:31

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$\newcommand{\a}{\alpha}\newcommand{\b}{\beta}$Consider the quadratic equation $$x^2 - (\a + \b)x + \a\b = 0.$$ It is easy to see that $\a$ and $\b$ both satisfy the equation. In particular, this means that $$x^2 = (\a + \b)x - \a\b$$ for $x = \a, \b$. Multiplying by $x^n$ for $n \geqslant 0$, we get $$x^{n + 2} = (\a + \b) x^{n + 1} - \a\b x^n.$$

The above is also satisfied by $\a$ and $\b$ both. Thus, adding the corresponding equations for $\a$ and $\b$, we get $$S_{n + 2} = (\a + \b) S_{n + 1} - \a\b S_n$$ for all $n \geqslant 0$. (With the convention that $S_0 := 2$.)

This technique similarly gives you the recursion for $\a^n - \b^n$ as well. I leave that to you.

Aryaman Maithani
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    The notations in the OP, $a,b,c$ for the coefficients of some $f(x)$, which should be $f(x)=ax^2+bx+c$ but is displayed in a way that should forbid an answer first, and then $\alpha,\beta$ for the roots, yes, the notations are not so good. But to go from $a,b,c;\alpha,\beta$ to $a,b,c;a,b$ is worse... – dan_fulea Jul 10 '21 at 11:28
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    @dan_fulea: I've changed $a, b$ to $\alpha, \beta$. – Aryaman Maithani Jul 10 '21 at 12:05
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    Yes, this is much better, thanks for the edit! – dan_fulea Jul 10 '21 at 12:13
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    What are the "corresponding equations for $\alpha$ and $\beta$"? – Peter Phipps Jul 10 '21 at 14:07
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    @PeterPhipps: The equations you get by plugging $x$ as $\alpha$ and $\beta$. More explicitly: $\newcommand{\a}{\alpha}\newcommand{\b}{\beta}$ $$\a^{n + 2} = (\a + \b) \a^{n + 1} - \a\b \a^n, \ \b^{n + 2} = (\a + \b) \b^{n + 1} - \a\b \b^n.$$ – Aryaman Maithani Jul 10 '21 at 14:10
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    @AryamanMaithani is there a difference b/w $\geqslant$ and $\geq$ or are they same? Thanks – p_square Jul 28 '21 at 16:25
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    @Algebrology: They're the same mathematically, these days I prefer $\geqslant$ for (purely) aesthetic reasons. :-) – Aryaman Maithani Jul 28 '21 at 18:03
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    @AryamanMaithani Thanks for your reply :) – p_square Jul 28 '21 at 18:08