Prove that:
$a^3\cos(B-C) + b^3\cos(C-A) + c^3\cos(A-B) = 3abc$
Here a,b,c are the sides of the triangle ABC opposite to angles A, B, C respectively.
My attempt: I tried putting $a = 2R \sin(A)$ and replacing $\sin(B+C) = \sin(A)$ but that didn't help out. I tried applying AM $\ge$ GM but that too did not work out for this question.
Please help me out!!!
$$\sum a^3\cos(B-C)$$
$$=\sum 2Ra^2\sin(A)\cos(B-C)$$
$$=\sum Ra^2(\sin(A+B-C)+\sin(A-B+C))$$
$$=\sum Ra^2(\sin(2C)+\sin(2B))$$
$$=\sum a^2 c\cos C + \sum a^2b\cos B$$
$$=\sum a^2c\cos C + \sum c^2a\cos A$$
$$=\sum ac(a\cos C+c\cos A)$$
$$=\sum abc=\fbox{3abc}$$