Can Master Theorem be applied on any of these? 1) $T(n) = 6T(n/2) + 2^{3 \log(n)}$
2) $T(n) = 8T(n/2) + \frac{n^3}{(\log(n))^4}$
3) $T(n) = 9T(n/3) + n(\log(n))^3$
Can the complexity for these be calculated with the Master Theorem?
I am not sure how to decide upon which case they fit in.
 A: You may be interested to know that there are exact solutions to your three problems. We will do problem one and provide references for the other two.
Suppose we have $T(0)$ and for $n\ge 1$ we have the recurrence
$$T(n) = 6 T(\lfloor n/2 \rfloor) + 2^{3\lfloor \log_2 n \rfloor}.$$
We can unroll this recurrence to obtain the exact result that
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor} 6^j \times 
2^{3(\lfloor \log_2 n \rfloor - j)}.$$
This simplifies to
$$T(n) = 2^{3\lfloor \log_2 n \rfloor} 
\sum_{j=0}^{\lfloor \log_2 n \rfloor} \left(\frac{6}{8}\right)^j
= 2^{3\lfloor \log_2 n \rfloor} 
\frac{1-(3/4)^{\lfloor \log_2 n \rfloor +1}}{1-3/4}
\\ =4 \times 2^{3\lfloor \log_2 n \rfloor} 
\left(1-(3/4)^{\lfloor \log_2 n \rfloor +1}\right).$$ 
The conclusion is that
$$T(n) \in \Theta\left(2^{3\lfloor \log_2 n \rfloor}\right)
= \Theta(n^3).$$
This MSE link points to a series of similar calculations.
A: For your second problem, lets solve the recurrence
$$T(n) = 8 T(\lfloor n/2 \rfloor) + 
\frac{n^3}{(1+\lfloor\log_2 n\rfloor)^4}$$
where $T(0) = 0.$
Let $$n = \sum_{k=0}^{\lfloor\log_2 n\rfloor} d_k 2^k$$ be the binary representation of $n.$
Unrolling the recursion we find the exact formula
$$T(n) = \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} d_k 2^{k-j}\right)^3.$$
Now to get an upper bound on this consider the case of $n$ being a string of one digits, which gives
$$T(n)\le \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} 2^{k-j}\right)^3
\\ = \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} 2^k\right)^3
=  \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4} (2^{\lfloor\log_2 n\rfloor+1} - 2^j)^3
\\ = \sum_{j=1}^{\lfloor\log_2 n\rfloor+1}
\frac{1}{j^4} (2^{\lfloor\log_2 n\rfloor+1} - 2^{\lfloor\log_2 n\rfloor+1-j})^3
= 2^{3(\lfloor\log_2 n\rfloor+1)}
\sum_{j=1}^{\lfloor\log_2 n\rfloor+1}
\frac{1}{j^4} (1 - 2^{-j})^3.$$
The sum term converges rapidly to a constant and hence the asymptotics of the upper bound are
$$\frac{1}{90}
\left(\pi^4 -270 \mathrm{Li}_4(1/2) + 270 \mathrm{Li}_4(1/4) - 90 \mathrm{Li}_4(1/8)\right)
\times 2^{3(\lfloor\log_2 n\rfloor+1)}.$$
For the a lower bound consider the case of a one digit followed by zeros, which gives
$$T(n)\ge \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(2^{\lfloor\log_2 n\rfloor-j}\right)^3
= 2^{3\lfloor\log_2 n\rfloor}\sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4}
\\ = 2^{3\lfloor\log_2 n\rfloor}\sum_{j=1}^{\lfloor\log_2 n\rfloor+1} \frac{1}{j^4} .$$
The sum term once again converges to a constant and the asymptotics of the lower bound are
$$\frac{\pi^4}{90} 2^{3\lfloor\log_2 n\rfloor}.$$
Joining the two bounds we see that
$$T(n) \in \Theta\left(2^{3\lfloor\log_2 n\rfloor}\right) = \Theta(n^3).$$
A: For problem three, lets solve the recurrence
$$T(n) = 9 T(\lfloor n/3\rfloor) + n \times (1+\lfloor\log_3 n\rfloor)^3$$
where we set $T(0)=0.$
Let the base three representation of $n$ be
$$n = \sum_{k=0}^{\lfloor\log_3 n\rfloor} d_k 3^k.$$
Then we get the exact formula
$$T(n) = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 9^j \times 
(1+\lfloor\log_3 n\rfloor- j)^3 \times 
\sum_{k=j}^{\lfloor\log_3 n\rfloor} d_k 3^{k-j}
\\ = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times 
(1+\lfloor\log_3 n\rfloor - j)^3 \times 
\sum_{k=j}^{\lfloor\log_3 n\rfloor} d_k 3^k .$$
Now for an upper bound consider a string of two digits, which gives
$$T(n)\le \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times 
(1+\lfloor\log_3 n\rfloor - j)^3 \times 2 \times
\sum_{k=j}^{\lfloor\log_3 n\rfloor} 3^k
\\ =  \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times 
(1+\lfloor\log_3 n\rfloor - j)^3 \left(3^{\lfloor\log_3 n\rfloor + 1}-3^j\right)
\\ = \sum_{j=1}^{\lfloor\log_3 n\rfloor+1} 
3^{\lfloor\log_3 n\rfloor+1-j} \times j^3 \times
\left(3^{\lfloor\log_3 n\rfloor + 1}-3^{\lfloor\log_3 n\rfloor + 1- j}\right)
\\ = 3^{2(\lfloor\log_3 n\rfloor + 1)}
\sum_{j=1}^{\lfloor\log_3 n\rfloor+1} 3^{-j} j^3 (1- 3^{-j}).$$
Now the sum term converges to a constant and we get the upper bound
$$\frac{7917}{2048} \times 3^{2(\lfloor\log_3 n\rfloor + 1)}.$$
For the lower bound take a one followed by a string of zeros, which gives
$$T(n)\ge \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times 
(1+\lfloor\log_3 n\rfloor - j)^3 \times 3^{\lfloor\log_3 n\rfloor}
\\ = 3^{\lfloor\log_3 n\rfloor} 
\sum_{j=1}^{\lfloor\log_3 n\rfloor+1} 3^{\lfloor\log_3 n\rfloor +1 -j} \times 
j^3
=  3^{2\lfloor\log_3 n\rfloor + 1} \sum_{j=1}^{\lfloor\log_3 n\rfloor+1} j^3 3^{-j}.$$
The sum term again converges to a constant and we get for the lower bound asymptotics the formula
$$\frac{33}{8}  3^{2\lfloor\log_3 n\rfloor + 1} .$$
Note that for the upper bound we have $\lfloor\log_3 n\rfloor+1\to\log_3 n,$ so that it is in fact $$\frac{7917}{2048} n^2 \approx 3.86572265625\times n^2$$ but for the lower bound $\lfloor\log_3 n\rfloor = \log_3 n$, so that it is $$\frac{33}{8} \times 3 \times n^2 = \frac{99}{8} n^2 \approx 12.375 \times n^2.$$
Joining the two bounds we may conclude that $$T(n)\in 
\Theta\left(3^{2\lfloor\log_3 n\rfloor}\right) = \Theta(n^2).$$
A similar computation which is a bit simpler can be used to analyse the cost of Strassen matrix multiplication.
