Let S = ∑_ r=1 ^ 3 cosec((π)/(8) + (r-1)(π)/(4)) cosec((π)/(8) +...

The given sum is $S = \sum_{r=1}^{3} \dfrac{1}{\sin\left(\dfrac{\pi}{8} + (r-1)\dfrac{\pi}{4}\right) \sin\left(\dfrac{\pi}{8} + r\dfrac{\pi}{4}\right)}$.

The difference between the angles in the denominator is $\dfrac{\pi}{4}$. Multiply and divide the entire sum by $\sin\left(\dfrac{\pi}{4}\right)$:

$S = \dfrac{1}{\sin\left(\dfrac{\pi}{4}\right)} \sum_{r=1}^{3} \dfrac{\sin\left(\left(\dfrac{\pi}{8} + r\dfrac{\pi}{4}\right) - \left(\dfrac{\pi}{8} + (r-1)\dfrac{\pi}{4}\right)\right)}{\sin\left(\dfrac{\pi}{8} + (r-1)\dfrac{\pi}{4}\right) \sin\left(\dfrac{\pi}{8} + r\dfrac{\pi}{4}\right)}$

Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$, the sum becomes a difference of cotangents:

$S = \sqrt{2} \sum_{r=1}^{3} \left[ \cot\left(\dfrac{\pi}{8} + (r-1)\dfrac{\pi}{4}\right) - \cot\left(\dfrac{\pi}{8} + r\dfrac{\pi}{4}\right) \right]$

This is a telescoping series. Expanding it yields:

$S = \sqrt{2} \left[ \left(\cot\dfrac{\pi}{8} - \cot\dfrac{3\pi}{8}\right) + \left(\cot\dfrac{3\pi}{8} - \cot\dfrac{5\pi}{8}\right) + \left(\cot\dfrac{5\pi}{8} - \cot\dfrac{7\pi}{8}\right) \right]$

$S = \sqrt{2} \left[ \cot\left(\dfrac{\pi}{8}\right) - \cot\left(\dfrac{7\pi}{8}\right) \right]$

Since $\cot\left(\dfrac{7\pi}{8}\right) = \cot\left(\pi - \dfrac{\pi}{8}\right) = -\cot\left(\dfrac{\pi}{8}\right)$, we get:

$S = \sqrt{2} \left[ \cot\left(\dfrac{\pi}{8}\right) - \left(-\cot\left(\dfrac{\pi}{8}\right)\right) \right] = 2\sqrt{2} \cot\left(\dfrac{\pi}{8}\right)$

Using the standard value $\cot\left(\dfrac{\pi}{8}\right) = \sqrt{2} + 1$:

$S = 2\sqrt{2}(\sqrt{2} + 1) = 4 + 2\sqrt{2}$

Comparing this with $S = a + b\sqrt{2}$, we find $a = 4$ and $b = 2$.

The roots of the required quadratic equation are $(a+b) = 6$ and $(a-b) = 2$.

Sum of the roots $= 6 + 2 = 8$
Product of the roots $= 6 \times 2 = 12$

The quadratic equation is $x^2 - (\text{Sum})x + (\text{Product}) = 0$, which gives $x^2 - 8x + 12 = 0$.

Answer: $x^2 - 8x + 12 = 0$