$\underline{\textbf{Solution 1}}$ Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \sin{A}$. By the Law of Sines, we must have $CA = \sin{B}$ and $AB = \sin{C}$. Now let us analyze the given: \begin{align*} &\cos^2A + \cos^2B + 2\sin A\sin B\cos C\\ &= 1-\sin^2A + 1-\sin^2B + 2\sin A\sin B\cos C\\ &= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C) \end{align*} Applying the Law of Cosines to simplify this: \[= 2-\sin^2C\] Therefore: \[\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.\] Similarly, \[\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.\]Note that the desired value is equivalent to $2-\sin^2B$, which is $2-\sin^2(A+C)$. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\dfrac{111-4\sqrt{35}}{72}$. Thus, the answer is $111+4+35+72 = \boxed{222}$. $\underline{\textbf{Solution 2}}$ Let \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ ------ (1)}\\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \text{ ------ (2)}\\ \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &= x \text{ ------ (3)}\\ \end{align*} Adding (1) and (3) we get: \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A( \sin B \cos C + \sin C \cos B) = \frac{15}{8} + x\]or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C) = \frac{15}{8} + x\]or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x\]or \[\cos^2 B + \cos^2 C = x - \frac{1}{8} --- (4)\] Similarly adding (2) and (3) we get: \[\cos^2 A + \cos^2 B = x - \frac{4}{9} --- (5)\]Similarly adding (1) and (2) we get: \[\cos^2 A + \cos^2 C = \frac{14}{9} - \frac{1}{8} --- (6)\] And (4) - (5) gives: \[\cos^2 C - \cos^2 A = \frac{4}{9} - \frac{1}{8} --- (7)\] Now (6) - (7) gives: $\cos^2 A = \frac{5}{9}$ or $\cos A = \sqrt{\dfrac{5}{9}}$ and $\sin A = \frac{2}{3}$ so $\cos C$ is $\sqrt{\dfrac{7}{8}}$ and therefore $\sin C$ is $\sqrt{\dfrac{1}{8}}$ Now $\sin B = \sin(A+C)$ can be computed first and then $\cos^2 B$ is easily found. Thus $\cos^2 B$ and $\cos^2 C$ can be plugged into (4) above to give x = $\dfrac{111-4\sqrt{35}}{72}$. Hence the answer is = $111+4+35+72 = \boxed{222}$.