To The Sixth

Note : It is very possible to solve for the possible values of $a$ and $b$ , but this is not recommended for this problem, as it leads to lots of unnecessary arithmetic. The alternative strategy, which I will illustrate below, only solves for the values of $ab$ and $a + b$ , and these are sufficient for computing the value(s) of $a^6 + b^6$ . This can be done by rewriting the given system of equations as a system of two equations in terms of these two expressions.

To start, note that $\begin{aligned} \left(a - \frac{1}{b}\right)\left(b - \frac{1}{a}\right) &= \frac{(ab - 1)^2}{ab} = a^2 + ab + b^2 \\ \frac{1}{a} + \frac{1}{b} + \frac{1}{ab} & = \frac{a + b + 1}{ab} = a + b \end{aligned}$ To simplify notation, let $x = a + b$ and $y = ab$ . Then, by adding $ab$ to both sides of the first equation, we obtain $\frac{2a^2b^2 - 2ab + 1}{ab} = (a + b)^2$ We now substitute $x$ and $y$ to obtain $\begin{aligned} x^2 & = \frac{2y^2 - 2y + 1}{y} \\ x & = \frac{1}{y - 1} \end{aligned}$ where the second of these comes from the second equation. This is a system of two equations and two unknowns, and can be solved as follows. We can substitute the second equation into the first to obtain $\frac{1}{(y - 1)^2} = \frac{2y^2 - 2y + 1}{y}$ which simplifies to $-2y^4 + 6y^3 - 7y^2 + 5y - 1 = 0$ This polynomial factors as follows: $-(y^2 - y + 1)(2y^2 - 4y + 1) = 0$ so by solving for $y$ and obtaining the respective values of $x$ (through back substitution), we end up with the following four pairs of values: $\begin{aligned} (x,y) & = \left(\sqrt{2}, 1 + \frac{\sqrt{2}}{2}\right) \\ (x,y) & = \left(-\sqrt{2}, 1 - \frac{\sqrt{2}}{2}\right) \\ (x,y) & = \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i, \frac{1}{2} + \frac{\sqrt{3}}{2}i\right) \\ (x,y) & = \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \end{aligned}$ With these values, we now take advantage of the fact that $a^6 + b^6 = (a^3 + b^3)^2 - 2a^3b^3$ and $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ . By substituting variables $x$ and $y$ , we obtain the equation $a^6 + b^6 = (x^3 - 3xy)^2 - 2y^3$ We quickly see that using any of the values of $x$ or $y$ with non-zero imaginary components gives such a complex value to $a^6 + b^6$ (due to the $xy$ term), so we only consider the first two pairs of values. Computing $a^6 + b^6$ using these values gives $a^6 + b^6 = 6 \pm \frac{5}{\sqrt{2}}$ the larger of which (obtained using the first pair) clearly must have the positive sign. So, we have $p = 6$ , $q = 5$ , and $r = 2$ . Thus, our answer is $p + q + r = 6 + 5 + 2 = \boxed{13}$ .