$\mathrm{\Sigma }{a}^3\cos (B-C)=$

Solution:

We have,

$\begin{array}{l}\sum a^3\cos (B-C)\\ =\sum k^3{\sin }^{3}A\cos (B-C)\\ =k^3{\mathrm{\Sigma sin}}^{2}A\sin (B+C)\cos (B-C)\\ =\frac{k^3}{2}{\mathrm{\Sigma sin}}^{2}A(\sin 2B+\sin 2C)\\ =\frac{k^3}{2}\mathrm{\Sigma }\left[{\sin }^{2}A(\sin 2B+\sin 2C)\left.+{\sin }^{2}B(\sin 2C+\sin 2A)+{\sin }^{2}C(\sin 2A+\sin 2B)\right]\right.\end{array}$

$\begin{array}{l}=k^3\mathrm{\Sigma }\left[{\sin }^{2}A\sin B\cos B+{\sin }^{2}B\sin A\cos A\right]\\ =k^3\mathrm{\Sigma sin}A\sin B\sin (A+B)\\ =k^3[\sin A\sin B\sin C+\sin B\sin C\sin A+\sin C\sin A\sin B]\\ =3(k\sin A)(k\sin B)(k\sin C)=3abc\end{array}$