## Task A-4.3.

Let $a, b$, and $c$ be distinct integers and $P(x) = x^{3} + a x^{2} + b x + c$ be a polynomial such that $P(a) = a^{3}$ and $P(b) = b^{3}$. Determine $P(1)$.
## First solution.

By substituting $a$ and $b$ into the polynomial $P(x)$ and using the conditions, we get

$$
a^{3}+a \cdot a^{2}+b \cdot a+c=a^{3} \quad \text { and } \quad b^{3}+a \cdot b^{2}+b \cdot b+c=b^{3}
$$

By subtracting and rearranging, we obtain

$$
\begin{array}{cl}
a^{3}+a b=a b^{2}+b^{2} & 1 \text { point } \\
\left(a^{2}+a b+b\right)(a-b)=0 . & 1 \text { point }
\end{array}
$$

Since $a$ and $b$ are distinct, it follows that $a^{2}+a b+b=0$, or $a^{2}=-b(a+1)$.

Notice that $a+1 \neq 0$, because otherwise one side of the last equation would be zero, while the other would not. Therefore, we conclude that

$$
\begin{aligned}
b & =-\frac{a^{2}}{a+1} \\
& =1-a-\frac{1}{a+1}
\end{aligned}
$$

Since $a$ and $b$ are integers, the number $a+1$ must be a divisor of 1, i.e., $a=0$ or $a=-2$.

If $a=0$, we get $b=0$, which contradicts the condition $a \neq b$. Therefore, it must be that $a=-2$.

In this case, we get $b=4$, and from either of the first two equations, we get $c=16$.

By direct verification, we see that this triplet $(a, b, c)$ indeed satisfies the conditions of the problem, so we have

$$
P(x)=x^{3}-2 x^{2}+4 x+16
$$

or

$$
P(1)=19
$$