3. The system of equations
$$
\left\{\begin{array}{l}
a=(b+c+d)^{3}, \\
b=(a+c+d)^{3}, \\
c=(a+b+d)^{3}, \\
d=(a+b+c)^{3}
\end{array}\right.
$$

has $(\quad)$ real solutions.
(A) 1
(B) 2
(C) 3
(D) More than 3
3. C.

By symmetry, without loss of generality, assume $a \geqslant b$, then
$$
\begin{array}{l}
(b+c+d)^{3} \geqslant(a+c+d)^{3} \\
\Rightarrow b+c+d \geqslant a+c+d \Rightarrow b \geqslant a .
\end{array}
$$

Thus, $a=b$.
Similarly, $a=c, a=d$.
Therefore, $a=b=c=d$.
Hence, $a=(3 a)^{3} \Rightarrow a=0$ or $\pm \frac{\sqrt{3}}{9}$.
Therefore, the system of equations has 3 real solutions.