$T(1) = 1 , T(n) = 2T(n/2) + n^3$? Divide and conquer $T(1) = 1 , T(n) = 2T(n/2) + n^3$? Divide and conquer, need help, I dont know how to solve it?

There  is another  closely  related recurrence  that  admits an  exact
solution.  Suppose we  have  $T(0)=0$  and for  $n\ge  1$ (this  gives
$T(1)=1$)
$$T(n) = 2 T(\lfloor n/2 \rfloor) + n^3.$$
Furthermore let the base two representation of $n$ be
$$n = \sum_{k=0}^{\lfloor \log_3 n \rfloor} d_k 2^k.$$
Then we  can unroll the  recurrence to obtain the  following exact
formula for $n\ge 1$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor} 
2^j \left(
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}
\right)^3.$$
Now to get an upper bound consider a string of one digits which yields
$$T(n) \le \sum_{j=0}^{\lfloor \log_2 n \rfloor} 
2^j \left(
\sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j}
\right)^3.$$
Note that this bound is attained and cannot be improved. It simplifies
to
$$\frac{32}{3} 8^{\lfloor \log_2 n \rfloor}
- 24 \times 4^{\lfloor \log_2 n \rfloor}
+ \left(6 \lfloor \log_2 n \rfloor + \frac{40}{3}\right) 
\times 2^{\lfloor \log_2 n \rfloor} + 1.$$
The lower bound is for the case of a one digit followed by a string of
zeros and yields
$$T(n) \ge \sum_{j=0}^{\lfloor \log_2 n \rfloor} 
2^j \left(
2^{\lfloor \log_2 n \rfloor-j}
\right)^3.$$
It simplifies to
$$\frac{4}{3} 8^{\lfloor \log_2 n \rfloor} 
- \frac{1}{3} 2^{\lfloor \log_2 n \rfloor}$$
and this matches the results posted by the other contributors.

Joining the dominant terms of the upper and the lower bound we obtain
the asymptotics
$$2^{3 \lfloor \log_2 n \rfloor}
\in \Theta\left(2^{3 \log_2 n}\right) 
= \Theta\left(n^3\right).$$
These are both in agreement with what the Master theorem would produce.
This MSE link has a series of similar calculations.
