3. In $\triangle A B C$, $A B=c, B C=a, C A=b, a^{4}+b^{4}+c^{4}=2 c^{2}\left(a^{2}+b^{2}\right)$, if $\angle A=72^{\circ}$, then $\angle B=$ $\qquad$
Solve:
\[
\left(a^{2}+b^{2}-c^{2}\right)^{2}=a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}-2 c^{2}\left(a^{2}+b^{2}\right)=2 a^{2} b^{2},
\]
then 
\[
\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} = \pm \frac{\sqrt{2}}{2} \Rightarrow C=45^{\circ} \text{ or } 135^{\circ} \text{ (discard)}, \text{ so } B=63^{\circ}.
\]