Let's determine the fifth-degree polynomial $P(x)$ such that $P(x)+8$ is divisible by $(x+1)^{3}$, and $P(x)-8$ is divisible by $(x-1)^{3}$.
The divisibility of the polynomial $P(x)+8$ by $(x+1)^{3}$ means that there exists a polynomial whose product with $(x+1)^{3}$ is identical to $P(x)+8$, and its degree is equal to the difference of the degrees of the two polynomials, $5-3=2$. Let this polynomial be $A x^{2}+B x+C$, and similarly, the quotient of $P(x)-8$ and $(x-1)^{3}$ is the polynomial $K x^{2}+L x+M$. Then the coefficients $A, B, C$ and $K, L, M$ must be determined so that the following two polynomials are identical:

$$
\begin{aligned}
& (x-1)^{3}\left(A x^{2}+B x+C\right)-8=A x^{5}+(3 A+B) x^{4}+ \\
& \quad+(3 A+3 B+C) x^{3}+(A+3 B+3 C) x^{2}+(B+3 C) x+(C-8), \text { and } \\
& (x-1)^{3}\left(K x^{2}+L x+M\right)+8=K x^{5}-(3 K-L) x^{4}+ \\
& \quad+(3 K-3 L+M) x^{3}-(K-3 L+3 M) x^{2}-(L-3 M) x+(-M+8)
\end{aligned}
$$

Two polynomials are identical if and only if the coefficients of the same powers of $x$ are equal. From this condition, we get the following system of equations for the unknown coefficients $A, B, C, K, L, M$:

$$
\begin{aligned}
A & =K, \\
3 A+B & =-3 K+L \\
3 A+3 B+C & =3 K-3 L+M, \\
A+3 B+3 C & =-K+3 L-3 M, \\
B+3 C & =-L+3 M, \\
C-8 \quad & =-M+8
\end{aligned}
$$

These, together with (1), can be rearranged as:

$$
\begin{aligned}
B-L & =-6 A \\
3(B+L)+(C-M) & =0 \\
3(B-L)-3(C+M) & =-2 A, \\
(B+L)+3(C-M) & =0 \\
C+M & =16
\end{aligned}
$$

From (4a), using (2a), (6), and (1), we get $A=K=3, B-L=18$, and from (3a) and (5), by adding them:

$$
B+L=-(C-M)=0
$$

thus

$$
\begin{gathered}
B=-9, \quad L=9, \quad C=M=8, \quad \text { finally } \\
P(x)=3 x^{5}-10 x^{3}+15 x
\end{gathered}
$$

László Jereb (Sopron, István Széchenyi Gymnasium, 2nd grade)