### Let $a$ and $b$ be positive real numbers such that$\dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0$ Find the value of $\left(\dfrac{b}{a} + \dfrac{a}{b} \right)^2$.

$5$

Step by Step Explanation:
1. We are given,
$\dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0$
Let $x = \dfrac{b}{a}$
$\implies b = ax$
2. Now,
\begin{align} & \implies \dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0 \\ & \implies \dfrac{1}{a} - \dfrac{1}{ ax } - \dfrac{1}{ a + ax } = 0 \\ & \implies \dfrac{1}{a}\left(1 - \dfrac{1}{ x } - \dfrac{1}{ 1 + x } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x(1 + x) - (1 + x) - x }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x + x^2 - 1 - x - x }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x^2 - x - 1 }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{ x^2 - x - 1 }{ x(1 + x) } = 0 && (\because a > 0) \\ & \implies x^2 - x - 1 = 0 \\ & \implies x = \dfrac{ 1 \pm \sqrt{ 1 + 4 } }{ 2(1) } \\ & \implies x = \dfrac{ 1 \pm \sqrt{5} }{ 2 } \\ & \text{ Since a and b are positive, x > 0 } \\ & \implies x = \dfrac{ 1 + \sqrt{ 5 } }{ 2 } \end{align}
3. Now,
\begin{align} \left( \dfrac{b}{a} + \dfrac{a}{b} \right)& = \left(x + \dfrac{1}{x} \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 }{ 1 + \sqrt{ 5 } } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 }{ 1 + \sqrt{ 5 } } \times \dfrac{ 1 - \sqrt{ 5 } }{ 1 - \sqrt{ 5 } } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 - 2\sqrt{5} }{ -4 } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } - \dfrac{1}{2} + \dfrac{ \sqrt{5} }{ 2 } \right)^2 \\ & = \left( \sqrt{5} \right)^2 \\ & = 5 \end{align}
4. Hence, the value of $\left(\dfrac{b}{a} + \dfrac{a}{b} \right)^2$ is $5$.