Question

Show that in $\begin{array}{c}\Delta ABC\\ \sum a^3\cos \left(B-C\right)=3abc\end{array}$

Answer 1

L.H.S. $=\sum a^3\cos (B-C)$

$=\sum a^2\left(2R\sin A\right)\cos (B-C)$

$=R\sum a^2.\left[2\sin \right(B+C\left)\cos \right(B-C\left)\right]$

$=R\sum a^2(\sin 2B+\sin 2C)$

$=R\sum a^2(2\sin B\cos B+2\sin C\cos C)$

$=\sum \left[a^2\right(2R\sin B)\cos B+a^2(2R\sin C\left)\cos C\right]$

$=\sum (a^{2}b\cos B+a^{2}c\cos C)$

$=(a^{2}b\cos B+a^{2}C\cos C)+(b^{2}c\cos C+b^{2}a\cos A)+(c^{2}a\cos A+c^{2}b\cos B)$

$=ab(a\cos B+b\cos A)+bc(b\cos C+c\cos B)+ca(C\cos A+a\cos C)$

$=ab\left(C\right)+bc\left(a\right)+ca\left(b\right)$

$=3abc=$ R.H.S.