Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$. 
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.

Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 -  (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align} 
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?

What the hey, I'll toss in one more (there are plenty of ways to work with the binomials, and some of the other answers already given are ones I'd have tried first):
We need $$ \ a + b \ = \ 7 \ \ , \ \ a - b \ = \ \sqrt{\Delta} \ = \ \sqrt{41} \ \ , \ \ ab \ = \ 2  \ \ , $$ $$ a^2 + b^2 \ = \ (a + b)^2 - 2ab \ = \ 49 - 4 \ = \ 45 \ \ , $$
as already presented. We can write
$$ (a + b)^6 \ + \ (a - b)^6 \ \ = \ \ 2a^6 \ + \ 2b^6 \ + \ 30·(a^4b^2 \ + \ a^2b^4) $$ $$ = \ \ 2·(a^6 \ + \ b^6) \ + \ 30·(ab)^2·(a^2 \ + \ b^2) \ \ = \ \ 2·(a^6 \ + \ b^6) \ + \ 30·2^2·45 $$
$$ = \ \ 7^6 \ + \ 41^3 \  \ = \ \ 49^3 \ + \ 41^3 \ \ . $$
If we need to do this "by hand", we have $ \ 49^3  +  41^3 \ = \ (49 + 41)·(49^2 - 49·41 + 41^2) \ \ , $ for which we can use little "binomial tricks" to produce $ \ (50 - 1)^2 \ - \ (50 - 1)·(40 + 1) \ + \ (40 + 1)^2 $ $ = \ 2401 - 2009 + 1681 \ = \ 2073 \ \Rightarrow \ 49^3  +  41^3 \ = \ 90·2073 \ = \ 207,300 - 20,730 $ $ = \  186,570 \ \ . $  We also have $ \ 30·4·45 \ = \ 6·4·15^2 \ = \ 6·(4·225) \ =  \ 6·900 \ = \ 5400 \ \ . $
Thus,
$$ 2·(a^6 \ + \ b^6) \ \ = \ \ [ \ (a + b)^6 \ - \ (a - b)^6 \ ] \ - \ 30·(ab)^2·(a^2 \ + \ b^2) $$
$$ = \ \ 186,570 \ - \ 5400 \ \ = \ \ 181,170 \ \ \Rightarrow \ \ a^6 \ + \ b^6 \ \ = \ \ 90,585 \ \ . $$
[This is somewhere between what you started out doing and Dan's method.]
