Trigonometry- Triangles Let $A,B,C$ by the angles of a triangle.
Then how to prove that 
$$a^3cos(B-C) + b^3cos(C-A) +c^3cos(A-B) = 3abc$$
I divided both sides by $abc$ and then tried to open the cosine function but nothing worked. I also took the cube of sine in the triple angle identity but that too just made the function more complicated and yielded no result. 

Let $2R=k$
$a=k\sin A, b=k\sin B,c=k\sin C$ (sine rule)
$$ a^3 \cos (B-C)= a^2\cdot a\cos (B-C)=$$
$$= a^2\cdot k\sin A\cos (B-C)= ka^2\sin (180-(B+C))\cos (B-C)=$$
$$= ka^2\sin (B+C)\cos (B-C)=\frac{ka^2}{2} (\sin 2B+\sin 2C) = $$
$$=\frac{ka^2}{2} (2\sin B\cos B+2\sin C\cos C) = ka^2(\sin B\cos B+\sin C\cos C) = $$
$$= a^2((k\sin B)\cos B+(k\sin C)\cos C)) =$$$$= a^2 (b\cos B+c\cos C)$$
The original expression now looks like this:
$$a^2(b\cos B+c\cos C)+b^2(c\cos C+a\cos A)+c^2(a\cos A+b\cos B) = $$
$$= a^2b\cos B + a^2c\cos C+b^2c\cos C+b^2a\cos A+c^2a\cos A+c^2b\cos B$$
$$= 
a^2b\cos B + 
b^2a\cos A +
b^2c\cos C +
c^2b\cos B +
a^2c\cos C +
c^2a\cos A
$$
$$=ab(a\cos B+b\cos A)+bc(b\cos C+c\cos B)+ca(a\cos C+c\cos A) =$$

$$= abc+bca+cab = 3abc$$
