Is it possible to apply the Master Theorem when there are multiple recursive calls? I'm trying to use the Master Theorem to solve the recurrence
$$T(n) = 3T(n/3) + 4T(n/4) + n^3$$
The problem is that there are multiple recursive calls of different sizes, so the recurrence doesn't have the form
$$T(n) = aT(n/b) + f(n)$$
that the Master Theorem applies to. Is there any way to get my recurrence into that form?
 A: For the recurrence
$$
T(n) = cT\left(\frac na\right)+dT\left(\frac nb\right)+f(n)
$$
considering $n = a^pb^qu$, $\{a,b,p,q,u\}\in \mathbb{N}^+$ with $a,b$ relative primes and $u\not| \{a,b\}$, the maximal paths to extinction are obtained by considering $m=p=q$. In those circumstances, considering null initial conditions for clarity, we have
$$
T\left(a^mb^mu\right)= cT\left(a^{m-1}b^mu\right)+dT\left(a^mb^{m-1}u\right)+f(a^mb^mu)
$$
and after recursion
$$
T\left(a^mb^mu\right)=\sum_{j=1}^m\left(\sum_{k=1}^m \left(\begin{array}{c}2m-j-k\\ m-j\end{array}\right)c^{m-j}d^{m-k}f(a^jb^k u)\right)
$$
NOTE
For $c = a$ and $d = b$ with $f(n) = n^3$ considering $u = 1$ we have
$$
T\left(a^mb^m\right)=\sum_{j=1}^m\left(\sum_{k=1}^m \left(\begin{array}{c}2m-j-k\\ m-j\end{array}\right)a^{m-j}b^{m-k}a^{3j}b^{3k}\right)
$$
or
$$
T\left(a^mb^m\right)=a^{3m}b^{3m}\sum_{j=1}^m\left(\sum_{k=1}^m \frac{\left(\begin{array}{c}2m-j-k\\ m-j\end{array}\right)}{a^{2(m-j)}b^{2(m-k)}}\right)
$$
and for $a=3, b = 4$ we have
$$
\lim_{m\to\infty}\sum_{j=1}^m\left(\sum_{k=1}^m \frac{\left(\begin{array}{c}2m-j-k\\ m-j\end{array}\right)}{a^{2(m-j)}b^{2(m-k)}}\right)\approx 1.2100840336134455
$$
hence
$$
T(n)=\mathcal{O}(n^3)
$$
