Sum of sixth power of roots of $x^3-x-1=0$ Question:
Find the sum of sixth power of roots of the equation $x^3-x-1=0$
My First approach:
Let $S_i$ denote the sum of $i^{th}$ power of roots of the given equation.
Now, multiplying given equation by $x^3$ , putting each of the three roots and adding the three formed equations
we get, $S_6=S_4+S_3$ 
repeating same above procedure to obtain, 
$S_4=S_2+S_1$ , $S_3=S_1+1$ 
hence, $S_6=S_2+2S_1+1=2+0+1=3$ 
My Second Approach:
Let $a,b,c$ be the roots of $f(x)=x^3-x-1$
Clearly, $f^/(x)/f(x)=1/(x-a)+1/(x-b)
+1/(x-c)$
=$\sum(1/x+a/x^2+a^2/x^3+a^3/x^4+\cdots)=3/x+S_1/x^2+S_2/x^3+\cdots$
hence we get, $S_6=5$
$\rule{200px}{0.5px}$
I am getting wrong answer through First Approach, Please point out my mistake or post a better solution.
Thank You 
 A: Squaring $x^3=x+1$, we have
$$
x^6=x^2+2x+1
$$
Furthermore, we know that the coefficient of $x^2$ in $x^3-x-1$ is the negative of the sum of the roots, and the coefficient of $x$ is the sum of the products of pairs of distinct roots.
Therefore, we need to find the sum of the squares of the roots, which is
$$
\left(\sum_{k=1}^3x_k\right)^2-2\sum_{j\lt k}x_jx_k=0^2-2(-1)=2
$$
The sum of the roots is $0$.
The sum of $1$ is $3$.
Thus, the sum of the sixth power of the roots is $2+0+3=5$.
A: 
$\textbf{Warning}$: The following approach is not the most efficient way of going about this—indeed, I probably wouldn't do this myself if I actually needed the answer.  However, it has pedagogical value and presents a rare opportunity for me to share a cool bit of math.  Without further adieu...


Let $\{\alpha_k\}_{k=1}^3$ be the roots of $f(x) = x^3 - x - 1$.  Notice that $P(\alpha_1, \alpha_2, \alpha_3) = \alpha_1^6 + \alpha_2^6 + \alpha_3^6$ is a symmetric polynomial* in $3$ variables evaluated at the $\alpha_k$'s.  It is a theorem** that any symmetric polynomial can be written as a polynomial in the elementary symmetric polynomials.  
Let $\big\{e_k(x_1,x_2,x_3) \big\}_{k=1}^3$ be the collection of elementary symmetric polynomials in three variables.  The above is saying that there exists a polynomial $Q \in \mathbb{Z}[x_1, x_2, x_3]$ such that $P(x_1, x_2, x_3) = Q\Big(e_1(x_1,x_2,x_3), \ e_2(x_1,x_2,x_3), \ e_3(x_1,x_2,x_3)\Big)$.
Now this looks like a mess, but the the elementary symmetric polynomials—evaluated at the $\alpha_k$'s—appear$^\dagger$ as the (signed) coefficients of $f$!  Thus, if we can find the polynomial $Q$ above, then we can easily find the desired value $P(\alpha_1, \alpha_2, \alpha_3)$ because the right hand side of this evaluation  will consist merely of the addition and multiplication of simple, already-known integers (see how robjohn got a clean natural number in his post?).  To be explicit, we'll have:
$$\begin{align}
&  e_1(\alpha_1, \alpha_2, \alpha_3) \ = \ 0 \qquad \ \ \text{ (the } x^2 \text{ coefficient)}\\
& e_2(\alpha_1, \alpha_2, \alpha_3) \ = \ -1 \\
& e_3(\alpha_1, \alpha_2, \alpha_3) \ = \ 1
\end{align}$$
Finding $Q$ may or may not be computationally difficult!  It appears that Newton did find a recursion which writes symmetric polynomials of the form $\displaystyle \sum_{k=1}^n x_k^j$ for any $n, j \in \mathbb{N}$ in terms of elementary symmetric polynomials (of course, in this scenario, we have $n=3$ and $j=6$).  See here for details.

$$\underline{\textbf{Footnotes}} \\[0.5em]$$
$\text{*}$The adjective "symmetric" here alludes to the fact that these polynomials do not change if you swap around ("permute") the variables.  Indeed, $x_1^6\!+\! x_2^6 \!+\! x_3^6  =  x_3^6 \!+ \!x_1^6 \!+ \!x_2^6$, etc.  On the other hand, $f(x_1, x_2) = x_1x_2^2$ would not be a symmetric polynomial as $x_1x_2^2 \neq x_2x_1^2$.

$\text{**}$Called the "fundamental theorem of symmetric polynomials", this was originally discovered by Isaac Newton.  See this thread for further discussion and links to several proofs.  

$\mathbf{^\dagger}$For any $n \in \mathbb{N}$, a collection of $n\!+\!1$ elementary symmetric polynomials in $n$ variables is defined.  To determine this collection, first start with a generic, completely-factored, $n^\text{th}$-degree monic polynomial $\displaystyle p(x) = \prod_{k=1}^n (x - \alpha_k)$.  After expanding and grouping terms, we'll get $p(x) = e_0x^n - e_1x^{n-1} + e_2x^{n-2} - \cdots + (-1)^ne_n$.  Each coefficient $e_k$ is the $k^\text{th}$ elementary symmetric polynomial in $n$ variables—a polynomial in $\mathbb{Z}[x_1, x_2, \cdots, x_n]$—which has been evaluated at the roots $x_k = \alpha_k$.  For example, let's look at the $n=2$ case:
$$p(x) \ = \ (x-\alpha_1)(x-\alpha_2) \ = \ x^2 - (\alpha_1 + \alpha_2)x + \alpha_1\alpha_2$$
This reveals that, in two variables, $e_0(x_1, x_2) = 1$ and $e_1(x_1, x_2) = x_1 + x_2$ and $e_2(x_1,x_2) = x_1x_2$. 
In generality, we'll have:
$$\begin{align}
& e_0(x_1, \cdots, x_n) \ = \ 1 \\[0.5em]
& e_1(x_1, \cdots, x_n) \ = \ \sum_{k=1}^n x_k \\[0.5em]
& e_2(x_1, \cdots, x_n) \ = \ \sum_{1 \leq k \leq j \leq n} x_kx_j \\[0.5em]
& e_3(x_1, \cdots, x_n) \ = \ \sum_{1 \leq k \leq j \leq m \leq n} x_kx_jx_m \\[0.5em]
& \ \ \vdots \qquad \text{ this pattern continues until: } \\[0.5em]
& e_n(x_1, \cdots, x_n) \ = \ \prod_{k=1}^n x_k
\end{align}$$
A: 
$S_6=S_2+2S_1+1$ 

This is not correct. 
We have
$$S_6=S_2+2S_1+\color{red}{3}$$
since $$S_6=x_1^6+x_2^6+x_3^6=(x_1^2+2x_1+1)+(x_2^2+2x_2+1)+(x_3^2+2x_3+1)$$
A: Let $a$, $b$ and $c$ be roots of the equation, $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Id est, $u=0$, $v^2=-\frac{1}{3}$ and $w^3=1$.
Thus, $$a^6+b^6+c^6=(a^3+b^3+c^3)^2-2(a^3b^3+a^3c^3+b^3c^3)=$$
$$=(27u^3-27uv^2+3w^3)^2-2(27v^6-27uv^2w^3+3w^6)=3^2-2((-1)^3+3)=5.$$
Done!
