Let's determine the values of $A, B, C$ if

$$
\frac{(x+1)^{2}}{(x+2)\left(x^{2}+x+1\right)} \equiv \frac{A}{x+2}+\frac{B x+C}{x^{2}+x+1}
$$
I. Solution. For the equation to be identical, it must hold for any value of $x$. Let's eliminate the fractions:

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

If $x=-2, \quad(-2+1)^{2}=1=A(4-2+1)=3 A$; hence $A=\frac{1}{3}$.

$$
\begin{array}{ll}
x=0, & 1=A+2 C ; 2 C=1-\frac{1}{3}=\frac{2}{3} ; C=\frac{1}{3} \\
x=-1, & 0=A(1-1+1)+(-B+C)(-1+2) \\
& 0=A+C-B, \text { so } B=A+C=\frac{2}{3}
\end{array}
$$

Based on these:

$$
\frac{(x+1)^{2}}{(x+2)\left(x^{2}+x+1\right)} \equiv \frac{1}{3}\left[\frac{1}{x+2}+\frac{2 x+1}{x^{2}+x+1}\right]
$$

Chabada György (Bencés g. VI. o. Győ́r)

II. Solution. In the previous solution, we saw that

$$
\begin{gathered}
(x+1)^{2} \equiv A\left(x^{2}+x+1\right)+(B x+C)(x+2) \\
x^{2}+2 x+1 \equiv(A+B) x^{2}+(A+2 B+C) x+A+2 C .
\end{gathered}
$$

From the identity, it follows that the coefficients of equal powers of $x$ are equal, so

$$
\begin{gathered}
A+B=1 \ldots \\
A+2 B+C=2 \ldots \\
A+2 C=1 \ldots
\end{gathered}
$$

Subtracting the terms of (1) from (2) and (3), we get

$$
\begin{array}{r}
B+C=1 \ldots \\
2 C-B=0 \ldots
\end{array}
$$

Thus:

$$
3 C=1, \quad C=\frac{1}{3}, \quad B=2 C=\frac{2}{3}, \quad A=1-B=1-\frac{2}{3}=\frac{1}{3} .
$$