How to Prove $a^3 \cos (B-C) + b^3 \cos (C-A)+ c^3 \cos (A-B)=3abc$?
If $A,$ $B,$ and $C$ are the vertices of the Triangle ABC, and $a,$ $b$ and $c$ is the side opposite to the sides of the triangle.
You may use -
Projection Formula:
$$a=b \cos C + c \cos B$$
and Sine and Cosine Relation too.
My try
$$a^3 \cos (B - C)+ b^3 \cos (C-A) + c^3 \cos (A-B)=3abc$$ LHS = $$ 3(b \cos C + c \cos B)(a \cos C + c \cos A)(a \cos B + b \cos A)$$ $$ 3[ ab \cos B \cos ^2 C + abc \cos A \cos B \cos C +a^2c \cos ^2 B \cos C +ac^2 \cos A \cos ^2B ]+[b^2 \cos A \cos^2 C + b^2c \cos^2 A \cos A+abc \cos A \cos B \cos C + ac^2 \cos A \cos^2 B] $$
Now I can't solve it further because there seems no way to reduce that further.
Thanks :)
It can be proved with sine-rule and cosine rule but it is really ugly.
$${\rm LHS} - {\rm RHS} = \sum_{cyc} ( a^3\cos(B-C) - abc ) = \sum_{cyc}( a^3(\cos B\cos C + \sin B\sin C) - abc)$$ By sine rule,
$$a : b : c = \sin A : \sin B : \sin C \quad\implies\quad \begin{cases} a^3\sin B\sin C = abc\sin A^2\\ b^3\sin C\sin A = abc\sin B^2\\ c^3\sin A\sin B = abc\sin C^2 \end{cases}$$ This leads to
$$\begin{align}{\rm LHS} - {\rm RHS} &= \sum_{cyc}( a^3 \cos B \cos C - abc\cos^2 A)\\ &= \sum_{cyc}\left[a^3\left(\frac{a^2+c^2-b^2}{2ac}\right)\left(\frac{a^2+b^2-c^2}{2ab}\right) - abc\left(\frac{b^2+c^2-a^2}{2bc}\right)^2\right]\\ &= \frac{1}{4abc}\sum_{cyc}a^2(\underbrace{a^4 - (b^2-c^2)^2 - (b^2+c^2-a^2)^2}_{I}) \end{align} $$ Notice what's in the square bracket equals to $$\require{cancel}I =\cancel{a^4} - (b^2-c^2)^2 - (b^2+c^2)^2 + 2a^2(b^2+c^2) - \cancel{a^4} =2(a^2(b^2+c^2) - b^4-c^4)$$ We obtain
$$\begin{align}{\rm LHS} - {\rm RHS} &= \frac{1}{2abc}\sum_{cyc} a^4b^2 + \color{red}{a^4}\color{green}{c^2} - a^2 b^4 - \color{blue}{a^2}\color{magenta}{c^4}\\ &= \frac{1}{2abc}\sum_{cyc} a^4b^2 + \color{red}{b^4}\color{green}{a^2} - a^2 b^4 - \color{blue}{b^2}\color{magenta}{a^4}\\ &= 0 \end{align}$$