Solving $T(n) = 2T(n/3) + T(2n/3) + n^3$
You can use the Akra-Bazzi theorem as
$$g(x) = n^3 \in O(n^c), a_1=2, a_2 = 1, b_1 = \frac{1}{3}, b_2 = \frac{2}{3}$$
So, just we need to find $p$ such that:
$$2\left(\frac{1}{3}\right)^p + \left(\frac{2}{3}\right)^p = 1$$
You can find the computation for finding $p$ here. It is around $1.4$. Now, we can write:
$$T(n) \in \Theta\left(n^{p}\left(1 +‌ \int_{1}^{n}\frac{u^3}{u^{1+p}}du\right)\right)$$
To simplify the term, we can write:
$$
\int_{1}^{n}\frac{u^3}{u^{1+p}}du = \int_{1}^{n}u^{2-p}du = \frac{1}{3-p}u^{3-p} |_1^n = \frac{1}{3-p}(n^{3-p}-1)
$$
Now, by substituting this term in the main complexity term, we can conclude that $T(n) \in \Theta(n^3)$.