<s>[INST] 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 
 [/INST] 
$S_6=S_2+2S_1+1$ 

This is not correct. 
We have
$$S_6=S_2+2S_1+\color{red}{3}$$
since $$S_6=x_1^6+x_2^6+x_3^6=(x_1^2+2x_1+1)+(x_2^2+2x_2+1)+(x_3^2+2x_3+1)$$
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