<s>[INST] 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?
 [/INST] Why work so hard?  From the quadratic formula we have:
$\{ a,b \} = \left\{ \frac{7-\sqrt{41}}{2}, \frac{7+\sqrt{41}}{2} \right\}$ so $a^6 + b^6 = 90585$.

My goodness!!!  A downvote on a simple, clear, correct answer.
Well, if the approach demands that the solution be done "by hand"—which was NOT part of the problem statement!— then here:
$$a^6 + b^6 = \frac{1}{2^6} \left[ ((7 - \sqrt{41})^2 )^3 + ((7 + \sqrt{41})^2 )^3\right]$$
$$= \frac{1}{64} \left[ (90 - 14 \sqrt{41})^3 + (90 + 14 \sqrt{41})^3\right]$$
Now cancel the terms with odd powers of $(14 \sqrt{41})$ (because they have opposite signs in the pairs), leaving
$$\frac{1}{64} \left[ 2 \cdot 90^3 + 6 \cdot 90 \cdot 14^2 \cdot 41\right] = 90585$$
Satisfied?
</s>