Let's determine the numbers $A, B, C, D$ such that

$$
\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C x+D}{x^{2}+x+1}=\frac{1}{(x+1)^{2} \cdot\left(x^{2}+x+1\right)}
$$

holds for all $x$ different from $-1$.
I. Solution. If there exists the desired quadruple of numbers $A, B, C, D$, then the polynomial equality is also an identity, which we obtain from (1) by multiplying by the denominator on the right side (this denominator is indeed a common multiple of the denominators on the left side, so multiplication results in an integer expression):

$$
A(x+1)\left(x^{2}+x+1\right)+B\left(x^{2}+x+1\right)+(C x+D)(x+1)^{2}=1
$$

Rearranging by powers of $x$:

$$
(A+C) x^{3}+(2 A+B+2 C+D) x^{2}+(2 A+B+C+2 D) x+(A+B+D)=1
$$

Accordingly, the coefficients of the corresponding powers of $x$ on both sides must be equal, and the terms not containing $x$ must also be equal:

$$
\begin{aligned}
A+C & =0 \\
2 A+B+2 C+D & =0 \\
2 A+B+C+2 D & =0 \\
A+B+D & =1
\end{aligned}
$$

Subtract from (3) both twice (2) and (4):

$$
\begin{aligned}
& B+D=0 \\
& C-D=0
\end{aligned}
$$

From (5) and (6), $A=1$, so from (2) $C=-1$, from (7) $D=-1$, and finally from (6) $B=1$. Thus, the desired quadruple exists, and (1) becomes an identity:

$$
\frac{1}{(x+1)^{2}\left(x^{2}+x+1\right)}=\frac{1}{x+1}+\frac{1}{(x+1)^{2}}-\frac{x+1}{x^{2}+x+1}
$$

Varga Katalin (Veszprém, Lovassy L. g. II. o. t.)

Remark. We can also obtain a system of equations for the numbers $A, B, C, D$ from $\left(1^{\prime}\right)$ by substituting permissible (i.e., not equal to -1) numbers for $x$. It is easy to compute if we take small absolute value integers, let $x=0,1,2$ and -2 (four equations are needed):

$$
\begin{array}{r}
A+B+D=1 \\
6 A+3 B+4 C+4 D=1 \\
21 A+7 B+18 C+9 D=1 \\
-3 A+3 B-2 C+D=1
\end{array}
$$

Subtract from (8) four times (5), from (9) nine times (5), and from (10) (5):

$$
\begin{array}{llr} 
& 2 A-B+4 C=-3 \\
12 A-2 B+18 C=-8, & \text { i.e., } & 6 A-B+9 C=-4 \\
-4 A+2 B-2 C=0, & \text { i.e., } & -2 A+B-C=0
\end{array}
$$

Add (13) to (11) and (12): $3 C=-3$, so $C=-1$, and $4 A+8 C=4 A-8=-4, A=1$, so from (13) $B=C+2 A=1$ and from (5) $D=-1$.

Kövesi Gusztáv (Budapest, I. István g. II. o. t.) Faragó István (Budapest, Könyves Kálmán g. II. o. t.)

II. Solution. If (1) holds, then

$$
\frac{A}{x+1}+\frac{C x+D}{x^{2}+x+1}=\frac{1}{(x+1)^{2}\left(x^{2}+x+1\right)}-\frac{B}{(x+1)^{2}}=\frac{1-B\left(x^{2}+x+1\right)}{(x+1)^{2}\left(x^{2}+x+1\right)}
$$

For this to be true, the right side must become a polynomial when multiplied by $(x+1)\left(x^{2}+x+1\right)$, or it must be simplifiable by $x+1$. The numerator can be factored into $x+1$ and another (first-degree) polynomial if its value is 0 when $x=-1$, since then the $x+1$ factor will be 0. Thus, $1-B \cdot 1=0, B=1$. Indeed, with this, the right side of (14) becomes

$$
\frac{-x(x+1)}{(x+1)^{2}\left(x^{2}+x+1\right)}=\frac{-x}{(x+1)\left(x^{2}+x+1\right)}
$$

Similarly, moving the first term of (14) to the right side:

$$
\frac{C x+D}{x^{2}+x+1}=\frac{-x}{(x+1)\left(x^{2}+x+1\right)}-\frac{A}{x+1}=\frac{-x-A\left(x^{2}+x+1\right)}{(x+1)\left(x^{2}+x+1\right)}
$$

the numerator on the right side must also give 0 when $x=-1$, from which $A=1$. With this, (15) becomes

$$
\frac{C x+D}{x^{2}+x+1}=\frac{-\left(x^{2}+2 x+1\right)}{(x+1)\left(x^{2}+x+1\right)}=\frac{-x-1}{x^{2}+x+1}
$$

which holds for all permissible $x$ if $C=-1, D=-1$.

Remark. This solution also sheds light on the reason why it is possible to find suitable values for $A, B, C, D$. (The systems of equations obtained above and any other similar systems of equations could have been contradictory.)