3. Given the quadratic equation in $x$
$$
a^{2} x^{2}+b^{2} x+c^{2}=0
$$

the sum of the roots is the sum of the squares of the roots of the quadratic equation
$$
a x^{2}+b x+c=0
$$

Then the relationship between $a$, $b$, and $c$ is ( ).
(A) $a^{2}=b c$
(B) $b^{2}=a c$
(C) $c^{2}=a b$
(D) $a b c=1$
3.B.

First, we have $a \neq 0$. By Vieta's formulas, the sum of the roots of equation (1) is $-\frac{b^{2}}{a^{2}}$. The relationships between the roots of equation (2) are $x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}$. Therefore, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=\frac{b^{2}}{a^{2}}-\frac{2 c}{a}$.
From the problem, we get $-\frac{b^{2}}{a^{2}}=\frac{b^{2}}{a^{2}}-\frac{2 c}{a}$, simplifying to $b^{2}=a c$.