Also...
since the value must be the same for all nonzero real numbers a and b such that |a| ≠ |b|
then we can choose for instance a = 1 and b = 2 and evaluate the expression.
\(\phantom{=\quad}\left( \frac{b^2}{a^2} + \frac{a^2}{b^2} - 2 \right) \times \left( \frac{a + b}{b - a} + \frac{b - a}{a + b} \right) \times \left( \frac{\frac{1}{a^2} + \frac{1}{b^2}}{\frac{1}{b^2} - \frac{1}{a^2}} - \frac{\frac{1}{b^2} - \frac{1}{a^2}}{\frac{1}{a^2} + \frac{1}{b^2}} \right) \\~\\ {=\quad}\left( \frac{2^2}{1^2} + \frac{1^2}{2^2} - 2 \right) \times \left( \frac{1 + 2}{2 - 1} + \frac{2 - 1}{1 + 2} \right) \times \left( \frac{\frac{1}{1^2} + \frac{1}{2^2}}{\frac{1}{2^2} - \frac{1}{1^2}} - \frac{\frac{1}{2^2} - \frac{1}{1^2}}{\frac{1}{1^2} + \frac{1}{2^2}} \right)\\~\\ {=\quad}\left(\frac41+\frac14-2 \right) \times \left( \frac{3}{1} + \frac{1}{3} \right) \times \left( \frac{\frac11 + \frac{1}{4}}{\frac{1}{4} - \frac{1}{1}} - \frac{\frac{1}{4} - \frac{1}{1}}{\frac{1}{1} + \frac{1}{4}} \right)\\~\\ {=\quad}\left(\frac94 \right) \times \left( \frac{10}{3} \right) \times \left( \frac{\frac54 }{ - \frac34} - \frac{-\frac34}{\frac54} \right)\\~\\ {=\quad}\left(\frac94 \right) \times \left( \frac{10}{3} \right) \times \left( \frac{5 }{ - 3} - \frac{-3}{5} \right)\\~\\ {=\quad}\left(\frac94 \right) \times \left( \frac{10}{3} \right) \times \left( -\frac{16}{15}\right)\\~\\ {=\quad}\left(\frac11 \right) \times \left( \frac{2}{1} \right) \times \left( -\frac{4}{1}\right)\\~\\ {=\quad}-8\)
And to make triple-sure, here's what WolframAlpha's result.