<s>[INST] 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?
 [/INST] 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 $9$.
</s>