$$2R(\cos^2 A+\cos^2B +\cos^2C)$$ $$=2R(\frac{3+\cos 2A +\cos 2B +\cos 2C}{2})$$ $$=R(3+(2\cos C \cos (A-B))+2\cos^2C-1)$$
$$=R(2+2\cos C (\cos (A-B)+\cos C))$$
I tried solving it further, but I kept getting confused with signs. How should I finish it?
Left hand side can be written as $$\sum2R\sin A \cos A = R \sum \sin 2A$$
$\sum \sin 2A$ is: $$\sin 2A + \sin 2B - \sin( 2A + 2B ) \\ = 2\sin(A+B)\cos(A-B) -2\sin(A+B)\cos(A+B) \\ = 4\sin(A+B)\sin A \sin B \\ = 4 \sin A \sin B \sin C$$
Thus, $LHS \ne RHS$. You might have copied your question wrongly.
In the standard notation we obtain:$$\sum_{cyc}a\cos\alpha=\sum_{cyc}\frac{a(b^2+c^2-a^2)}{2bc}=\sum_{cyc}\frac{a^2b^2+a^2c^2-a^4}{2abc}=$$ $$=\sum_{cyc}\frac{2a^2b^2-a^4}{2abc}=\frac{16S^2}{2abc}=4\cdot\frac{abc}{4S}\cdot\frac{8S^3}{a^2b^2c^2}=4R\sin\alpha\sin\beta\sin\gamma$$
Let $O$ be the circumcenter of $ABC$. Then $d(O,AB)=R\cos C$ and $$ 2[ABC] = 2[OAB]+2[OBC]+2[OAC] $$ leads to $$ \frac{abc}{2R} = aR\cos A+bR\cos B+cR\cos C $$ where the LHS also equals $4R^2 \sin A\sin B\sin C$, so $$ a \cos A + b\cos B + c\cos C = \color{red}{4} R\sin A\sin B\sin C.$$
