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?

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

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

Can you finish?

• i am Trying by this…. Jun 28 at 9:27
• Yes, @DavidQuinn i found tha answer as 9. Thanks very much Jun 28 at 9:30

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 $$\sum_k x_k^3=12-3=9$$

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

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

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

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 $$9$$.

• Yes, thanks bud Jun 29 at 0:34
• Good luck buddy! Jun 29 at 4:44