9.16. Find the sum

$$
1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}
$$

## 9.4. Pairing



##
9.16. According to problem $9.121^{3}+2^{3}+3^{3}+\ldots+m^{3}=\left(\frac{m(m+1)}{2}\right)^{2}$. Therefore, $1^{3}+2^{3}+3^{3}+\ldots+(2 n-1)^{3}+(2 n)^{3}=\left(\frac{2 n(2 n+1)}{2}\right)^{2}$, i.e.
$1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}+2^{3}\left(1^{3}+2^{3}+\ldots+n^{3}\right)=\left(\frac{2 n(2 n+1)}{2}\right)^{2}$. Transform the last equality, using the fact that $1^{3}+$ $+2^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2} \cdot$ As a result, we get $1^{3}+3^{3}+5^{3}+\ldots$ $\cdots+(2 n-1)^{3}=n^{2}\left(2 n^{2}-1\right)$.