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?

$$P(x)=x^4-3x+1=0, \\ R(P(x))=0$$
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
$$\sum R^{3}(P(x))=-15.$$
