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?

From Vieta’s rules, $a+b=7, ab=2$. $$a^3+b^3=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)= 7(49-2\cdot 3)=7\cdot 43=301 $$ Again, $$(a^3+b^3)^2=a^6+b^6+2a^3b^3$$$$301^2=a^6+b^6+2(2)^3$$ so $$a^6+b^6=301^2-16=90585. $$
