3-ча 1. How many real solutions does the system of two equations with three unknowns have:

$$
\left\{\begin{aligned}
x+y & =2 \\
x y-z^{2} & =1 ?
\end{aligned}\right.
$$
Sol 1. From the second equation, it follows that $x y \geqslant 1$. The numbers $x$ and $y$ cannot both be negative, since their sum is 2. Therefore, the numbers $x$ and $y$ are positive and $x+y \geqslant 2 \sqrt{x y} \geqslant 2$, and the equality $x+y=2$ is possible only when $x=y=1$. In this case, $z=0$.

Part 2. Solve the system of equations:

$$
\left\{\begin{aligned}
x^{3}-y^{3} & =26 \\
x^{2} y-x y^{2} & =6
\end{aligned}\right.
$$

Sol 2. Let $y=k x$. Immediately note that $k \neq 1$. From the equations

$$
\begin{aligned}
x^{3}-k^{3} x^{3} & =26 \\
k x^{3} y-k^{2} x^{3} & =6
\end{aligned}
$$

we get $x^{3}=\frac{26}{1-k^{3}}$ and $x^{3}=\frac{6}{k-k^{2}}$. Therefore,

$$
\frac{26}{1-k^{3}}=\frac{6}{k-k^{2}}
$$

This equation can be multiplied by $1-k$. As a result, we get

$$
\frac{26}{1+k+k^{2}}=\frac{6}{k}
$$

from which $k=3$ or $\frac{1}{3}$. Therefore, $x^{3}=-1$ or $x^{3}=27$. In the end, we get the following solutions: $(-1,-3)$, $\left(\frac{1 \pm i \sqrt{3}}{2}, \frac{3}{2}(1 \pm i \sqrt{3})\right),(3,1),\left(\frac{-3 \pm 3 i \sqrt{3}}{2}, \frac{1}{2}(-1 \pm i \sqrt{3})\right)$.

Part 3. Find the sum

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

Sol 3. By induction on $m$, it is easy to prove that $1^{3}+2^{3}+3^{3}+\cdots+m^{3}=\left(\frac{m(m+1)}{2}\right)^{2}$. Indeed, the base of induction is obvious, so we only need to check the equality

$$
\frac{m^{2}(m+1)^{2}}{2}+(m+1)^{3}=\frac{(m+1)^{2}(m+2)^{2}}{.}
$$

After canceling by $m+1$ and multiplying by 4, we get the obvious equality $m^{2}+4(m+1)=(m+2)^{2 *}$

Thus, $1^{3}+2^{3}+3^{3}+\cdots+(2 n-1)^{3}+(2 n)^{3}=\left(\frac{2 n(2 n+1)}{2}\right)$, i.e., $1^{3}+3^{3}+5^{3}+\cdots+(2 n-1)^{3}+2^{3}\left(1^{3}+2^{3}+\cdots+n^{3}\right)=\left(\frac{2 n(2 n+1)}{2}\right)^{2}$. Transform the last equality, using the fact that $1^{3}+2^{3}+\cdots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$. As a result, we get $1^{3}+3^{3}+5^{3}+\cdots+(2 n-1)^{3}=n^{2}\left(2 n^{2}-1\right)$.

## Series C