106. Find the sum: $1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}$.
Solution. Let's denote the sum by $S_{n}$:

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

We need to find an expression for the general term $S_{n}$ that would allow us to compute the sum $S_{n}$ for any $n$ without raising all numbers from 1 to $2 n-1$ to the third power.

Consider the first differences of the sequence defined by formula (1). The general form of the first differences is

$$
\begin{gathered}
d_{n}=S_{n+1}-S_{n}=\left(1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}+(2 n+1)^{3}\right)- \\
-\left(1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}\right)=(2 n+1)^{3}
\end{gathered}
$$

- a polynomial of the third degree. Therefore, the general term of the sequence defined by formula (1) can be represented as a polynomial of the fourth order:

$$
S_{n}=a x^{4}+b x^{3}+c x^{1}+d x+l .
$$

To determine the coefficients $a, b, c, d$ and $l$, substitute the values of $S_{n}$ for $n=1,2,3,4,5$ into this formula, then we get the system of equations:

$$
\left\{\begin{aligned}
a+b+c+d+l & =1 \\
16 a+8 b+4 c+2 d+l & =28 \\
81 a+27 b+9 c+3 d+l & =153 \\
256 a+64 b+16 c+4 d+l & =496 \\
625 a+125 b+25 c+5 d+l & =1225
\end{aligned}\right.
$$

The solutions to this system are: $a=2, b=0, c=-1, d=l=0$.

Therefore, $S_{n}=2 n^{4}-n^{2}=n^{2}\left(2 n^{2}-1\right)$.