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 

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!
