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, $S_6=5$
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I am getting wrong answer through First Approach, Please point out my mistake or post a better solution.
Thank You 

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 $2+0+3=5$.
