3. Given non-zero real numbers $a, b, c$ satisfy
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {. }
$$

Then for positive integer $k$ such that
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}$,
(2) $\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}=\frac{1}{a^{2 k+1}+b^{2 k+1}+c^{2 k+1}}$,
(3) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k}=\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}$,
(4) $\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}=\frac{1}{a^{2 k}+b^{2 k}+c^{2 k}}$

Among them, the number of always true statements is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
3. B.

From the given equation, we have
$$
\begin{array}{l}
(a b+b c+c a)(a+b+c)=a b \cdot c \\
\Rightarrow(a+b)(b+c) \cdot(c+a)=0 .
\end{array}
$$

Thus, $a+b=0$ or $b+c=0$ or $c+a=0$.
By symmetry, assume $a+b=0$. Then
$$
\begin{array}{l}
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{c^{2 k+1}} \\
=\frac{1}{a^{2 k+1}+b^{2 k+1}+c^{2 k+1}} \\
=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}} .
\end{array}
$$

Therefore, equations (1) and (2) hold.