Finding $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$ in $\Delta \text{ABC}$ I am attempting the following trigonometry problem. Given that in an acute angled triangle $\Delta \text{ABC}$, the following equalities hold true

$$\cos^2(A)+\cos^2(B)+2\sin(A)\sin(B)\cos(C)=\dfrac{15}{8}\\
\cos^2(B)+\cos^2(C)+2\sin(B)\sin(C)\cos(A)=\dfrac{14}{9}$$
Find the value of $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$.


My Attempt:
Let the unknown quantity be $x$. Then we have, by adding all the terms.
$$2\sum_{cyc}\cos^2(A)-2\sum_{cyc}\sin(A)\sin(B)\cos(A+B)=\dfrac{15}{8}+\dfrac{14}{9}+x$$
Also simplifying the second summation term as follows, we get $$\sin(A)\sin(B)\cos(A+B)=\dfrac{\sin(2A)\sin(2B)}{4}-(1-\cos^2(A))(1-\cos^2(B))$$

I'm not sure how to proceed further. Any hints are appreciated. Even hints to other possible pathways to the solution are welcome. Thanks

Hint: Note that $$2\sin A\sin B\cos C-2\cos A\cos B\cos C=2\cos(180^\circ-A-B)\cdot\cos C=2\cos^2 C$$
So, the first equation becomes $$\cos^2 A+\cos^2 B+\cos^2 C+2\cos A\cos B\cos C=\frac{15}8$$This is symmetric, which makes it quite useful.
Now, we subtract the second equation from this to get $$\cos^2 A+2\cos A\cos B\cos C-2\sin B\sin C\cos A=\frac{15}8-\frac{14}9$$$$=\cos^2 A+2\cos(B+C)\cos A$$You can take it from here!
