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

$$
\frac{(x+3)^{2}}{(x-2)\left(x^{2}+x+1\right)}=\frac{A}{x-2}+\frac{B x+C}{x^{2}+x+1}
$$

is an identity.
I. solution. Of course, only those $x$ can be in question which do not make any denominator zero, otherwise the expressions would be meaningless. The only value to be excluded is $x=2$, because $x^{2}+x+1=(x+0.5)^{2}+0.75$, which is always positive. Thus, multiplying by the left-hand side denominator,

$$
(x+3)^{2}=A\left(x^{2}+x+1\right)+(B x+C)(x-2)
$$

is also an identity. Reducing to zero and rearranging:

$$
(A+B-1) x^{2}+(A-2 B+C-6) x+(A-2 C-9)=0
$$

This can only hold for all $x$ different from 2 if each coefficient is individually 0, because otherwise the left side would be a quadratic or linear polynomial, or a non-zero constant, which could only be zero at two, or one place, respectively, not at all. The requirement of the problem can therefore only be met, and is then obviously met, if

$$
\begin{gathered}
A+B-1=0 \\
A-2 B+C-6=0 \\
A-2 C-9=0
\end{gathered}
$$

Add to (3) twice (2) and half of (4):

$$
3.5 A-12.5=0, \quad \text { from which } \quad A=25 / 7
$$

From this, using (2) and (4), $B=-18 / 7, C=-19 / 7$, so the sought identity is:

$$
\frac{(x+3)^{2}}{(x-2)\left(x^{2}+x+1\right)} \equiv \frac{25}{7(x-2)}-\frac{18 x+19}{7\left(x^{2}+x+1\right)}
$$

Károly Ambrus (Budapest, I. László g. II. o. t.)

II. solution. The fact that (1) is an identity means that it holds for all $x$ for which the expressions on both sides are defined. Thus, by choosing an arbitrary value for $x$, we get an equation for $A, B, C$. Since we have 3 unknowns, we choose 3 values, preferably ones that make the calculation easy. Let them be $x=0,1$ and -1, with these

$$
9=A-2 C, \quad 16=3 A-B-C, \quad 4=A+3 B-3 C
$$

from which the above triplet of values for $A, B, C$ again follows.

Tamás Görbe (Budapest, Bem J. g. II. o. t.)

Remarks. 1. It is not by chance that the equation obtained by substituting $x=0$ is identical to (4), since in (4) we also only considered the terms not containing $x$.

2. The line of thought in the first solution shows that if the equality in the problem is an identity, then (1) holds for all $x$, including $x=2$, so in the second solution, this could also have been used. This would have been advantageous, because then we would have immediately obtained the equation $25=7 A$ with one unknown for $A$.