Question

Show that:

$\sum a^3\cos (B-C)=3abc$

(or)

Prove that:

$a^3\cos (B-C)+b^3\cos (C-A)+c^3\cos (A-B)=3abc$

Answer 1

To prove: $\sum a^3\cos (B-C)=a^3\cos (B-C)+b^3\cos (C-A)+c^3\cos (A-B)=3abc$

Recall the Sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Where, $R$ is the circumradius of the triangle.

$\Rightarrow a=2R\sin A,b=2R\sin B$ and $c=2R\sin C$

Now, lets take LHS and then equate it to RHS.

LHS $=\sum a^3\cos (B-C)$

$=\sum a^{2}a\cos (B-C)$

$=\sum a^2\left(2R\sin A\right)\cos (B-C)$---(1)

Since, in $\Delta ABC$

$A+B+C={180}^{\circ }$

$\Rightarrow A={180}^{\circ }-(B+C)$

$\Rightarrow \sin A=\sin ({180}^{\circ }-(B+C\left)\right)$

$\Rightarrow \sin A=\sin (B+C)$

Now, from (1)

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

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

$=\sum a^{2}R\left(\sin \right(B+C+B-C)+\sin (B+C-B+C))$

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

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

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

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

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

By cyclic expansion of $a,b$ and $c$

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

$=ab(a\cos B+b\cos A)+ac(a\cos C+c\cos A)+bc\left(b\cos C\right)c\cos B)$

$=ab\left(c\right)+ac\left(b\right)+bc\left(a\right)$

$=3abc$

$=$ RHS

Therefore, $\sum a^3\cos (B-C)=3abc$

Hence, proved.