\( \Large x + \frac{a}{x} = b => \frac{x^{2}+a}{x} = b \)

=> \( \Large x^{2} + a = bx \) ...(i)

Now, \( \Large \frac{x^{2}+bx+a}{bx^{2}-x^{3}} => \frac{ \left(x^{2}+a\right) + bx} {bx^{2}-x^{3}} \) [using Eq. (i)]

= \( \Large \frac{2bx}{bx^{2} - x^{3}} = \frac{2b}{bx - x^{2}} = \frac{2b}{a} \)

\( \Large Because, x^{2} + bx \)

\( \Large bx - x^{2} = a \)