In $\triangle A B C$, $$ a^3 \cdot \cos (B-C)+b^3 \cdot \cos (C-A)+c^3 \cdot \cos (A-B)= $$
In $\triangle A B C$,
$$
a^3 \cdot \cos (B-C)+b^3 \cdot \cos (C-A)+c^3 \cdot \cos (A-B)=
$$
a b c
a + b + c
2 abc
3 abc
Solution
In $\begin{aligned} & \triangle A B C, a^3 \cos (B-C) \\ & =a^3\left(\frac{2 \sin (B+C) \cos (B-C)}{2 \sin (B+C)}\right) \\ & =a^3\left(\frac{\sin 2 B+\sin 2 C)}{2 \sin (B+C)}\right) \\ & =a^3\left(\frac{2 \sin B \cos B+2 \sin c \cos C}{2 \sin (\pi-A)}\right) \\ & =a^3\left(\frac{\sin B \cos B+\sin C \cos C}{\sin A}\right) \\ & =a^3\left(\frac{b k \cos B+c K \cos C}{a k}\right)\end{aligned}$
Adding Eqs. (i), (ii) and (iii), we get
$
\begin{aligned}
& a^3 \cos (B-C)+b^3 \cos (C-A)+c^3 \cos (A-B) \\
& =a^2 b \cos B+a^2 c \cos C+b^2 c \cos C \\
& \quad+b^2 a \cos A+c^2 a \cos A+c^2 b \cos B \\
& =a b(a \cos B+b \cos A)+a c(a \cos C+c \cos A) \\
& =a b c+a b c+a b c=3 a b c \quad+b c(b \cos C+c \cos B)
\end{aligned}
$
