T(n) = aT(n/b) + f(n)
You applied the version when f(n) = O(n^(log_b(a) - e)) for some e > 0.
This is important, you need this to be true for some e > 0.
For f(n) = n^3, b = 2 and a = 8,
n^3 = O(n^(3-e)) is not true for any e > 0.
So your picked the wrong version of the Master theorem.
You need to apply a different version of Master theorem:
if f(n) = Theta ((log n)^k * n^log_b(a)) for some k >= 0,
then
T(n) = Theta((log n)^(k+1) * n^log_b(a))
In your problem, you can apply this case, and that gives T(n) = Theta(n^3 log n).
An alternative way to solve your problem would be:
T(n) = 8 T(n/2) + n^3.
Let g(n) = T(n)/n^3.
Then
n^3 *g(n) = 8 * (n/2)^3 * g(n/2)+ n^3
i.e g(n) = g(n/2) + 1.
This implies g(n) = Theta(logn) and so T(n) = Theta(n^3 logn).