4. Find the sum $1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}$.
4. Use the formula from problem 3.

$$
1^{3}+2^{3}+3^{3}+\ldots+(2 n)^{3}=\left[\frac{2 n(2 n+1)}{2}\right]^{2}
$$

Isolate the sum of cubes of odd numbers

$$
\begin{gathered}
1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}+2^{3}+4^{3}+\ldots+(2 n)^{3}=\left[\frac{2 n(2 n+1)}{2}\right]^{2} \\
S+2^{3}\left(1^{3}+2^{3}+\ldots+n^{3}\right)=\left[\frac{2 n(2 n+1)}{2}\right]^{2}, \quad S+8\left[\frac{n(n+1)}{2}\right]^{2}=\left[\frac{2 n(2 n+1)}{2}\right]^{2}
\end{gathered}
$$

Answer: $S=n^{2}(2 n+1)^{2}-2 n^{2}(n+1)^{2}=n^{2}\left(2 n^{2}-1\right)$.

$6,7,8,9,10,11$. The validity of the specified formulas can be proven by mathematical induction.