## Task 1.

(a) Determine all ordered triples $(a, b, c)$ of real numbers for which

$$
a+b+c=2021 \quad \text { and } \quad a^{2} b+b^{2} c+c^{2} a=a^{2} c+b^{2} a+c^{2} b
$$

(b) Determine all ordered triples $(a, b, c)$ of real numbers for which

$$
a+b+c=2021 \quad \text { and } \quad a^{3} b+b^{3} c+c^{3} a=a^{3} c+b^{3} a+c^{3} b
$$
## Solution.

(a) The condition

$$
a^{2} b+b^{2} c+c^{2} a=a^{2} c+b^{2} a+c^{2} b
$$

is successively equivalent to

$$
\begin{gathered}
a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=0 \\
a^{2}(b-c)+b^{2}(c-b+b-a)+c^{2}(a-b)=0 \\
a^{2}(b-c)+b^{2}(c-b)+b^{2}(b-a)+c^{2}(a-b)=0 \\
a^{2}(b-c)+b^{2}(c-b)+b^{2}(b-a)+c^{2}(a-b)=0 \\
\left(a^{2}-b^{2}\right)(b-c)+\left(c^{2}-b^{2}\right)(a-b)=0 \\
(a-b)(a+b)(b-c)+(c-b)(c+b)(a-b)=0 \\
(a-b)(b-c)(a+b-c-b)=0 \\
(a-b)(b-c)(a-c)=0
\end{gathered}
$$

From this, we conclude that at least two of the numbers $a, b, c$ must be equal.

Using the condition $a+b+c=2021$, we get that all solutions are given by

$$
\begin{aligned}
& (a, b, c)=(t, t, 2021-2 t), t \in \mathbb{R} \\
& (a, b, c)=(t, 2021-2 t, t), t \in \mathbb{R} \\
& (a, b, c)=(2021-2 t, t, t), t \in \mathbb{R}
\end{aligned}
$$

(b) The condition

$$
a^{3} b+b^{3} c+c^{3} a=a^{3} c+b^{3} a+c^{3} b
$$

is successively equivalent to

$$
\begin{gathered}
a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=0 \\
a^{3}(b-c)+b^{3}(c-b+b-a)+c^{3}(a-b)=0 \\
a^{3}(b-c)+b^{3}(c-b)+b^{3}(b-a)+c^{3}(a-b)=0
\end{gathered}
$$

$$
\begin{gathered}
a^{3}(b-c)+b^{3}(c-b)+b^{3}(b-a)+c^{3}(a-b)=0 \\
\left(a^{3}-b^{3}\right)(b-c)+\left(c^{3}-b^{3}\right)(a-b)=0 \\
(a-b)\left(a^{2}+a b+b^{2}\right)(b-c)+(c-b)\left(c^{2}+b c+b^{2}\right)(a-b)=0 \\
(a-b)(b-c)\left(a^{2}+a b+b^{2}-c^{2}-b c-b^{2}\right)=0 \\
\left.(a-b)(b-c)\left(a^{2}-c^{2}+a b-b c\right)\right)=0 \\
(a-b)(b-c)((a-c)(a+c)+b(a-c))=0 \\
(a-b)(b-c)(a-c)(a+c+b)=0
\end{gathered}
$$

or, using the condition $a+b+c=2021$, we have

$$
2021(a-b)(b-c)(a-c)=0
$$

We can then conclude as in part (a) of the problem.