If a and b are nonzero real numbers such that \(\left| a \right| \ne \left| b \right|\), compute the value of the expression\( \left( \frac{b^2}{a^2} + \frac{a^2}{b^2} - 2 \right) \times \left( \frac{a + b}{b - a} + \frac{b - a}{a + b} \right) \times \left( \frac{\frac{1}{a^2} + \frac{1}{b^2}}{\frac{1}{b^2} - \frac{1}{a^2}} - \frac{\frac{1}{b^2} - \frac{1}{a^2}}{\frac{1}{a^2} + \frac{1}{b^2}} \right). \)
+128473 First term simplified = ( b^4 + a^4 - 2a^2b^2) (a^2 - b^2)^2
___________________ = ___________
a^2b^2 a^2b^2
Second term simplified = [(a + b)^2 + (b - a)^2 ] 2a^2 + 2b^2 2 (a^2 + b^2)
___________________ = ____________ = __________
b^2 - a^2 - (a^2 - b^2) - (a^2 - b^2)
Third term simplified = b^2 + a^2 a^2 - b^2
_________ - ___________ =
a^2 - b^2 a^2 + b^2
(a^2 + b^2)^2 - (a^2 - b^2)^2
________________________ =
(a^2 - b^2) (a^2 + b^2)
[ (a^2 + b^2) + (a^2 - b^2] * [ (a^2 + b^2) - ( a^2 - b^2) ] ( 2a^2) ( 2b^2)
____________________________________________ = __________________ =
(a^2 - b^2) ( a^2 + b^2) (a^2 - b^2)(a^2 + b^2)
4a^2b^2
__________________
(a^2 - b^2) (a^2 + b^2)
Mutiplying the first term by the third we have
(a^2 - b^2)^2 4a^2b^2 4(a^2 - b^2)
____________ * _____________________ = _____________
a^2b^2 (a^2 - b^2) (a^2 + b^2) (a^2 + b^2)
Multiply this result by the second term and we have
4(a^2 - b^2) 2(a^2 + b^2) 4 * 2
___________ * ____________ = ______ = - 8
(a^2 + b^2) - (a^2 - b^2) -1
+9466 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. 
