FindthevalueofC

Given that:

$\begin{aligned} 4a^2 - 5ab + b^2 & = 0 & \small \blue{\text{Divide both sides by }ab} \\ 4\left(\frac ab\right) - 5 + \frac ba & = 0 & \small \blue{\text{Let }x = \frac ab} \\ 4x - 5 + \frac 1x & = 0 & \small \blue{\text{Multiply both sides by }x} \\ 4x^2 - 5a + 1 & = 0 \\ (4x-1)(x-1) & = 0 \\ \implies x & = \frac 14, \ 1 \end{aligned}$

Then we have:

$\begin{aligned} \frac {ab+8a^2}{4a^2-b^2} & = \frac {1+8x}{4x-\frac 1x} & \small \blue{\text{Divide up and down by }ab} \\ & = \begin{cases} \dfrac {1+2}{1-4} = - 1 & \small \text{for }x = \dfrac 14 \\ \dfrac {1+8}{4-1} = 3 & \small \text{for }x = 1 \end{cases} \end{aligned}$

Therefore the answer is $\boxed{\text{3 and -1}}$ .

Upon observation, $4a^2 - 5ab + b^2 = (4a-b)(a-b) = 0 \Rightarrow b = a, 4a.$ Substituting these values of $b$ yields $\boxed{C = 3, -1}.$