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If a and b are nonzero real numbers such that \(\left| a \right| \ne \left| b \right|\), compute the value of the expression\( \left( \frac{b^2}{a^2} + \frac{a^2}{b^2} - 2 \right) \times \left( \frac{a + b}{b - a} + \frac{b - a}{a + b} \right) \times \left( \frac{\frac{1}{a^2} + \frac{1}{b^2}}{\frac{1}{b^2} - \frac{1}{a^2}} - \frac{\frac{1}{b^2} - \frac{1}{a^2}}{\frac{1}{a^2} + \frac{1}{b^2}} \right). \)

 Jul 30, 2019
 #1
avatar+103148 
+2

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

 

 

cool cool cool

 Jul 30, 2019
 #2
avatar+8725 
+3

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 resultsmiley

 Jul 30, 2019
 #3
avatar+103148 
+1

LOL, hectictar....eveyone knows that "three of a kind" is a tough hand to beat......!!!!!

 

 

 

cool cool cool

CPhill  Jul 30, 2019

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