5. It is known that one of the roots of the equation $x^{2}-4 a^{2} b^{2} x=4$ is $x_{1}=\left(a^{2}+b^{2}\right)^{2}$. Find $a^{4}-b^{4}$. Answer. $\pm 2$.
Solution. Substituting $x=x_{1}$ into the given equation, we get $\left(a^{2}+b^{2}\right)^{4}-4 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}=4$, from which $\left(a^{2}+b^{2}\right)^{2}\left(\left(a^{2}+b^{2}\right)^{2}-4 a^{2} b^{2}\right)=4,\left(a^{2}+b^{2}\right)^{2}\left(a^{2}-b^{2}\right)^{2}=4,\left(a^{4}-b^{4}\right)^{2}=4$, therefore, $a^{4}-b^{4}= \pm 2$.