Question

$a^3\cos (B-C)+b^3\cos (C-A)+c^3\cos (A-B)$ is equal to

• A. $3abc$
• B. $3(a+b+c)$
• C. $abc(a+b+c)$
• D. $0$

Answer 1

Answer: (A)

$3abc$

$a^3\cos (B-C)=a^2\left(a\cos (B-C)\right)$
Using sine rule for $a=k\sin A$, we get
$\Rightarrow a^3\cos (B-C)=a^2(ksinA.\cos (B-C))$
We have $A+B+C=\pi$ and $A=\pi -(B+C)$
$\Rightarrow \sin A=\sin (\pi -(B+C))=\sin (B+C)$
$\therefore a^3\cos (B-C)=a^{2}k\sin (B+C)\cos (B-C)$
$=a^2\frac{k}{2}(\sin 2B+\sin 2C)$
$=a^2\frac{k}{2}(2\sin B\cos B+2\sin C\cos C)$
$=a^2(k\sin B\cos B+k\sin C\cos C)$
$=a^2(b\cos B+c\cos C)$ ..............$\left(1\right)$
Now,$b^3\cos (C-A)=b^{2}k\sin (C+A)\cos (C-A)$
$=b^2\frac{k}{2}(\sin 2C+\sin 2A)$
$=b^2\frac{k}{2}(2\sin C\cos C+2\sin A\cos A)$
$=b^2(k\sin C\cos C+k\sin A\cos A)$
$=b^2(c\cos C+a\cos A)$ ..............$\left(2\right)$
Again,$b^3\cos (C-A)=c^{2}k\sin (A+B)\cos (A-B)$
$=c^2\frac{k}{2}(\sin 2A+\sin 2B)$
$=c^2\frac{k}{2}(2\sin A\cos A+2\sin B\cos B)$
$=c^2(k\sin A\cos A+k\sin B\cos B)$
$=c^2(a\cos A+b\cos B)$ ..............$\left(3\right)$
Adding eqns$\left(1\right),\left(2\right)$ and $\left(3\right)$ we get
$a^3\cos (B-C)+b^3\cos (C-A)+c^3\cos (A-B)$
$=a^2(b\cos B+c\cos C)+b^2(c\cos C+a\cos A)+c^2(a\cos A+b\cos B)$
Taking common, we get
$=ab(a\cos B+b\cos A)+bc(b\cos C+c\cos B)+ca(c\cos A+a\cos C)$
We know that $a\cos B+b\cos A=c,b\cos C+c\cos B=a$ and $c\cos A+a\cos C=b$
$=abc+abc+abc=3abc$