Find the value of $α^3+β^3+γ^3+δ^3$ given that $α,β,γ,δ$ are roots of $x^4+3x+1=0$ 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?
 A: $$x^4-3x+1=0\implies x^3=3-\frac{1}{x}$$
This is satisfied by each of the roots, so $$\Sigma x^3=\Sigma3-\Sigma\frac{1}{x}$$
Can you finish?
A: $$P(x)=x^4-3x+1=0, \\ R(P(x))=0$$
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
$$\sum R^{3}(P(x))=-15.$$
A: 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
$$\sum_k x_k^3=12-3=9$$
A: 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 $9$.
