A strange equation

From the equation $\frac {1}{a} - \frac {1}{b} = \frac {1}{a+b}$ , multiplying throughout by $a+b$ gives $\frac {a+b} {a} - \frac {a+b}{b} = 1$ , and so $\frac {b}{a} - \frac {a}{b} = 1$ . This gives $\left( \frac {b}{a} + \frac {a}{b}\right)^2 = \left(\frac {b}{a} - \frac {a}{b}\right)^2 + 4 \times \frac {b}{a} \times \frac {a}{b} = 1 + 4 = 5$ . Since $a$ and $b$ are positive, we take the positive square root to obtain $\frac {b}{a} + \frac {a}{b} = \sqrt{5}$ .

Taking the third power, we get

$5 \sqrt{5} = \left(\frac {b}{a} + \frac {a}{b}\right)^3 = \frac {b^3}{a^3} + 3 \frac {b}{a} + 3 \frac {a}{b} + \frac {a^3} {b^3},$

implying $\frac {a^3}{b^3} + \frac {a^3} {b^3} = 2 \sqrt{5}$ . Finally, squaring gives $20 = \left(\frac {a^3}{b^3} + \frac {b^3} {a^3} \right)^2 = \frac {a^6}{b^6} + 2 + \frac {b^6}{a^6}$ , so $\frac {a^6}{b^6} + \frac {b^6} {a^6} = 18$ .

$\frac { 1 }{ a } -\frac { 1 }{ b } =\frac { 1 }{ a+b } \left( a\neq 0,b\neq 0,a+b\neq 0 \right) \\ \frac { a+b }{ a } -\frac { a+b }{ b } =1\quad (multiplying\quad by\quad a+b)\\ \Longrightarrow \quad 1+\frac { b }{ a } -\left\{ \frac { a }{ b } +1 \right\} =1\\ \Longrightarrow \quad \frac { b }{ a } -\frac { a }{ b } =1\\ \Longrightarrow { \left( \frac { b }{ a } -\frac { a }{ b } \right) }^{ 3 }=1\Longrightarrow \left( \frac { b }{ a } \right) ^{ 3 }-\quad { \left( \frac { a }{ b } \right) }^{ 3 }-3\left( \frac { b }{ a } -\frac { a }{ b } \right) =1\Longrightarrow \left( \frac { b }{ a } \right) ^{ 3 }-\quad { \left( \frac { a }{ b } \right) }^{ 3 }=4\\ \therefore { \left[ \left( \frac { b }{ a } \right) ^{ 3 }-\quad { \left( \frac { a }{ b } \right) }^{ 3 } \right] }^{ 2 }=16\Longrightarrow \left( \frac { b }{ a } \right) ^{ 6 }+\quad { \left( \frac { a }{ b } \right) }^{ 6 }-\quad 2\quad =\quad 16\\ \therefore \boxed { \left( \frac { a }{ b } \right) ^{ 6 }+\quad { \left( \frac { b }{ a } \right) }^{ 6 }=\quad 18 }$