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Let $α,β,γ,δ$ be the roots(real or non real) of the equation $x^4-3x+1=0$. Then find the value of $α^3+β^3+γ^3+δ^3$.
I tried this question as $S_1=0, S_2=0, S_3=3, S_4=1$ and then I used $S_1^3$ to find the the value of asked question, but i am not able to factorise it further and it seems like a dead end. Moreover it is very lengthy method so can you tell of any other more elegant way of approaching this question?

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$$x^4-3x+1=0\implies x^3=3-\frac{1}{x}$$

This is satisfied by each of the roots, so $$\Sigma x^3=\Sigma3-\Sigma\frac{1}{x}$$

Can you finish?

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  • $\begingroup$ i am Trying by this…. $\endgroup$ Jun 28 at 9:27
  • $\begingroup$ Yes, @DavidQuinn i found tha answer as 9. Thanks very much $\endgroup$ Jun 28 at 9:30
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As by @David Quinn if roots are $x_k$, then $$\sum_kx^3_k=3\sum_{k=1}^4 1-\sum_k\frac{1}{x_k}=12-\sum_k y_k~~~~(1)$$ now transform the $x$ equation by $x=1/y$, then $y^4-3y^3+1=0 \implies \sum_k y_k=3~~~~(2)$ Eq. (1) and (2) give $$\sum_k x_k^3=12-3=9$$

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$$P(x)=x^4-3x+1=0, \\ R(P(x))=0$$

Let

  • $R(P(x))$ be a root of $P(x)$

  • $R^{-1}(P(x))$ be an inverse root of $P(x)$

  • $R^3(P(x))=\left(R(P(x))\right)^3.$

$$x^3=-3-\frac 1x$$

$$\begin{align}\sum R^{-1}(P(x))=-12-\sum R^3(P(x))\end{align}$$

Applying, $$-\frac{P^{'}(0)}{P(0)} =\sum R^{-1}(P(x))$$

We get

$$\sum R^{3}(P(x))=-15.$$

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Using Vieta's theorem for $x^4+3x+1=0$, we get: $$a+b+c+d=0 \Rightarrow b+c+d=-a\\ ab+ac+ad+bc+bd+cd=0 \Rightarrow bc+bd+cd=a^2\\ abc+abd+acd+bcd=-3\Rightarrow a^3=-bcd-3\\ abcd=1$$ Hence: $$a^3+b^3+c^3+d^3=(-bcd-3)+(-acd-3)+(-abd-3)+(-abc-3)=\\ -(abc+abd+acd+bcd)-12=3-12=-9.$$ Note: $x^4+3x+1=0$ is given in the title, but $x^4-3x+1=0$ is given in the body. The same method for the second equation will produce $9$.

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  • $\begingroup$ Yes, thanks bud $\endgroup$ Jun 29 at 0:34
  • $\begingroup$ Good luck buddy! $\endgroup$
    – farruhota
    Jun 29 at 4:44

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