## Task A-1.3.

For real numbers $x_{1}, x_{2}, \ldots, x_{30}$, the following holds:

$$
\begin{aligned}
& 20^{3} x_{1}+21^{3} x_{2}+\cdots+49^{3} x_{30}=13 \\
& 21^{3} x_{1}+22^{3} x_{2}+\cdots+50^{3} x_{30}=1 \\
& 22^{3} x_{1}+23^{3} x_{2}+\cdots+51^{3} x_{30}=19
\end{aligned}
$$

What is the value of $21 x_{1}+22 x_{2}+\cdots+50 x_{30}$?
## First solution.

It is known that for real numbers $a$ and $b$, $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$.

If $a-b=1$, then $a^{3}-b^{3}=a^{2}+a b+b^{2}$. Therefore, by subtracting the first equation from the second, we get:

$\left(21^{2}+21 \cdot 20+20^{2}\right) x_{1}+\left(22^{2}+22 \cdot 21+21^{2}\right) x_{2}+\cdots+\left(50^{2}+50 \cdot 49+49^{2}\right) x_{30}=-12$.

Similarly, by subtracting the second equation from the third, we get:

$$
\left(22^{2}+22 \cdot 21+21^{2}\right) x_{1}+\left(23^{2}+23 \cdot 22+22^{2}\right) x_{2}+\cdots+\left(51^{2}+51 \cdot 50+50^{2}\right) x_{30}=18 . \quad 2 \text { points }
$$

Notice that for any real number $a$:

$$
\left((a+1)^{2}+(a+1) \cdot a+a^{2}\right)-\left(a^{2}+a \cdot(a-1)+(a-1)^{2}\right)=6 a
$$

Therefore, by subtracting the two previously obtained equations, we get:

$$
6 \cdot\left(21 x_{1}+22 x_{2}+\cdots+50 x_{30}\right)=30
$$

from which $21 x_{1}+22 x_{2}+\cdots+50 x_{30}=5$.