Question

In $△ABC$,
$a^3\cdot \cos (B-C)+b^3\cdot \cos (C-A)+c^3\cdot \cos (A-B)=$

• A. a b c
• B. a+b+c
• C. 2 a b c
• D. 3 a b c

Answer 1

Answer: (D)

In $\Delta ABC,a^3\cos (B-C)$
$=a^3\left(\frac{2\sin (B+C)\cos (B-C)}{2\sin (B+C)}\right)$
$=a^3\left(\frac{\sin 2B+\sin 2C)}{2\sin (B+C)}\right)$
$=a^3\left(\frac{2\sin B\cos B+2\sin c\cos C}{2\sin (\pi -A)}\right)$
$=a^3\left(\frac{\sin B\cos B+\sin c\cos C}{\sin A}\right)$
$=a^3\left(\frac{bk\cos B+cK\cos C}{ak}\right)$
$\therefore a^3\cos (B-C)=a^{2}b\cos B+a^{2}c\cos C...(i)$
Similarly,
$b^3\cos (C-A)=b^{2}c\cos C+b^{2}a\cos A...(ii)$
and $c^3\cos (A-B)=c^{2}a\cos A+c^{2}b\cos B...(iii)$
Adding Eqs. (i), (ii) and (iii), we get
$a^3\cos (B-C)+b^3\cos (C-A)+c^3\cos (A-B)$
$=a^{2}b\cos B+a^{2}c\cos C+b^{2}c\cos C+b^{2}a\cos A+c^{2}a\cos A+c^{2}b\cos B$
$=ab(a\cos B+b\cos A)+ac(a\cos C+c\cos A)+bc(b\cos C+c\cos B)$
$=abc+abc+abc=3abc$