Finding closed form for $1^3+3^3+5^3+...+n^3$

By factorizing $2^3 = 8$ from the even terms, we can re-express the sum of the even terms:

$$2^3 + 4^3 + \dots + (n-1)^3 = 8\left(1^3 + 2^3 + \dots + \left(\frac{n-1}{2}\right)^3\right)$$

Simply subtract this from $1^3 + 2^3 + \dots + n^3$ to give the sum of the remaining odd terms. You can compute the formula for the RHS of the above equation using the equation you mentioned.

It may be simpler to use a method that does not require any creativity:

Let n = (2k-1) here.

$\sum_{i=1}^k(2i-1)^3 = \sum_{i=1}^k(8i^3 - 12i^2 + 6i - 1) = 8\sum_{i=1}^ki^3 - 12\sum_{i=1}^ki^2 + 6\sum_{i=1}^ki - \sum_{i=1}^k1 = 8(\frac{k(k+1)}{2})^2 - 12\frac{k(k+1)(2k+1)}{6} + 6\frac{k(k+1)}{2} - k$

Using repeated differences we get $$ \begin{array}{llll} 1 & 28 & 153 & 496 & 1225 & 2556 & \\ 27 & 125 & 343 & 729 & 1331 & \\ 98 & 218 & 386 & 602 & \\ 120 & 168 & 216 & \\ 48 & 48 & \\ 0 & \\ \end{array} $$ Newton's interpolation formula then gives us $$ 1 \binom{n-1}{0} + 27 \binom{n-1}{1} + 98 \binom{n-1}{2} + 120 \binom{n-1}{3} + 48\binom{n-1}{4} = 2 n^4-n^2 $$

This is a technique well worth knowing. (The simplification was done with WA.)