In triangle $\mathrm{ABC}$, the following relationship holds between the sides:

$$
\left(a^{2}+b^{2}+c^{2}\right)^{2}=4 b^{2}\left(a^{2}+c^{2}\right)+3 a^{2} c^{2}
$$

What is the measure of angle $\beta$?
From both sides of (1), take the first term on the right side! The left side can then be transformed as follows:

$$
\begin{aligned}
\left(a^{2}+b^{2}+c^{2}\right)^{2}-4 b^{2}\left(a^{2}+c^{2}\right) & =\left(a^{2}+c^{2}\right)^{2}+b^{4}+2 b^{2}\left(a^{2}+c^{2}\right)-4 b^{2}\left(a^{2}+c^{2}\right)= \\
& =\left(a^{2}+c^{2}-b^{2}\right)^{2}
\end{aligned}
$$

Thus, (1) holds exactly when

$$
\left(a^{2}+c^{2}-b^{2}\right)^{2}=3 a^{2} c^{2}
$$

Let's write the cosine rule for the $\beta$ angle of triangle $ABC$, which is opposite to side $b$:

$$
\begin{gathered}
b^{2}=a^{2}+c^{2}-2 a c \cos \beta, \quad \text { hence } \\
a^{2}+c^{2}-b^{2}=2 a c \cos \beta
\end{gathered}
$$

Substituting this into (2) gives

$$
(2 a c \cos \beta)^{2}=4 a^{2} c^{2} \cos ^{2} \beta=3 a^{2} c^{2}
$$

from which

$$
\cos ^{2} \beta=\frac{3}{4}, \quad \text { since } \quad a c \neq 0
$$

Thus, $\cos \beta=\frac{\sqrt{3}}{2}$ or $\cos \beta=-\frac{\sqrt{3}}{2}$. In the first case, $\beta=30^{\circ}$, and in the second case, $\beta=150^{\circ}$. Therefore, the $\beta$ angle of the triangle is either $30^{\circ}$ or $150^{\circ}$. Since the steps of the solution are reversible, it is also clear that if a triangle has a $\beta$ angle of $30^{\circ}$ or $150^{\circ}$, then for its sides - with the usual notation - (1) holds.