On the recurrence $T(n) = T(n/a) + T(n/b) + n^c$
Suppose that $$ T(n) = T(n/a) + T(n/b) + n^c. $$
Let's try to prove by induction that $T(n) \leq Mn^C$:
$$
T(n) \leq M(n/a)^c + M(n/b)^c + n^c = \left((a^{-c} + b^{-c})M + 1 \right) n^C.
$$
Therefore we need $(a^{-c} + b^{-c})M + 1 \leq M$, which we can satisfy as long as $a^{-c} + b^{-c} &lt; 1$.
When $a^{-c} + b^{-c} = 1$, the answer is likely $T(n) = \Theta(n^c\log n)$.