Functions

Im just going to brute force it. You may find other answers.

$f(f(f(x))) = 64x - 105 - 63x + 278$

$f(f(px+q)) = 64x - 105 - 63x + 278$

$f(p^2x+pq+q) = 64x - 105 - 63x + 278$

$p^3x+p^2q+pq+q = 64x - 105 - 63x + 278$

From this, we get that 

$p^3 = 64 - 63 = 1$

$p = 1$

Therefore, 

$3q = 173$

$q = 57\frac{2}{3}$

So, $p+q=58\frac{2}{3}$