Question

In $\triangle A B C$,
$a^{3} \cdot \cos (B-C)+b^{3} \cdot \cos (C-A)+c^{3} \cdot \cos (A-B)=$

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

Answer 1

Answer: (D)

In $\Delta A B C, a^{3} \cos (B-C)$
$=a^{3}\left(\frac{2 \sin (B+C) \cos (B-C)}{2 \sin (B+C)}\right)$
$=a^{3}\left(\frac{\sin 2 B+\sin 2 C)}{2 \sin (B+C)}\right)$
$=a^{3}\left(\frac{2 \sin B \cos B+2 \sin c \cos C}{2 \sin (\pi-A)}\right)$
$=a^{3}\left(\frac{\sin B \cos B+\sin c \cos C}{\sin A}\right)$
$=a^{3}\left(\frac{b k \cos B+c K \cos C}{a k}\right)$
$\therefore a^{3} \cos (B-C)=a^{2} b \cos B+a^{2} c \cos C\,\,\,...(i)$
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
$a^{3} \cos (B-C)+b^{3} \cos (C-A)+c^{3} \cos (A-B)$
$=a^{2} b \cos B+a^{2} c \cos C+b^{2} c \cos C+b^{2} a \cos A+c^{2} a \cos A+c^{2} b \cos B$
$=a b(a \cos B+b \cos A)+a c(a \cos C+c \cos A)+b c(b \cos C+c \cos B)$
$ =a b c+a b c+a b c=3 a b c$