Find all quadruples of real numbers $(a, b, c, d)$ satisfying the system of equations

$$
\left\{\begin{array}{l}
(b+c+d)^{2010}=3 a \\
(a+c+d)^{2010}=3 b \\
(a+b+d)^{2010}=3 c \\
(a+b+c)^{2010}=3 d
\end{array}\right.
$$
There are two solutions: $(0,0,0,0)$ and $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.

If $(a, b, c, d)$ satisfies the equations, then we may as well assume $a \leq b \leq c \leq d$. These are non-negative because an even power of a real number is always non-negative. It follows that

$$
b+c+d \geq a+c+d \geq a+b+d \geq a+b+c
$$

and since $x \mapsto x^{2010}$ is increasing for $x \geq 0$ we have that

$$
3 a=(b+c+d)^{2010} \geq(a+c+d)^{2010} \geq(a+b+d)^{2010} \geq(a+b+c)^{2010}=3 d
$$

We conclude that $a=b=c=d$ and all the equations take the form $(3 a)^{2010}=3 a$, so $a=0$ or $3 a=1$. Finally, it is clear that $a=b=c=d=0$ and $a=b=c=d=\frac{1}{3}$ solve the system.