Example 4 (53rd Romanian Mathematical Olympiad (Final)) Find all real numbers $a, b, c, d, e \in[-2,2]$, such that $a+b+c+d+e=0, a^{3}+b^{3}+c^{3}+d^{3}+e^{3}=0, a^{5}+b^{5}+c^{5}+d^{5}+e^{5}=10$.


Let's perform a variable substitution, set $a=2 \cos x, b=2 \cos y, c=2 \cos z, d=2 \cos t, e=2 \cos u$. Since $2 \cos 5 x=$ $(2 \cos x)^{5}-5(2 \cos x)^{3}+5(2 \cos x)=a^{5}-5 a^{3}+5 a$, therefore, denoting $\sum$ as the cyclic sum,
$\sum 2 \cos 5 x=\sum a^{5}-5 \sum a^{3}+5 \sum a=10$, which means $\sum \cos 5 x=5$.
Thus, $\cos 5 x=\cos 5 y=\cos 5 z=\cos 5 t=\cos 5 u=1$.
From $\cos 5 x=1$, we get $5 x=2 k \pi, k \in \mathbf{Z}$.
Since $\cos 0=1, \cos \frac{2 \pi}{5}=\frac{\sqrt{5}-1}{4}, \cos \frac{4 \pi}{5}=-\frac{\sqrt{5}+1}{4}$, therefore, $a, b, c, d, e \in\left\{2, \frac{\sqrt{5}-1}{2},-\frac{\sqrt{5}+1}{2}\right\}$.
From $\sum a=0$ we know that these 5 numbers must be 1 of 2, 2 of $\frac{\sqrt{5}-1}{2}$, and 2 of $-\frac{\sqrt{5}+1}{2}$.
The above $a, b, c, d, e$ satisfy $\sum a^{3}=0$.