Question

In triangle $ABC,$ let $a,b,c$ be the lengths of the sides opposite to the angles $A,B,C$ respectively. Let the value of $a^3\cos 3B+3a^{2}b\cos (A-2B)+3a{b}^2\cos (2A-B)+b^3\cos 3A$ be $l.$ Then the value of $\frac{l}{c^3}$ is equal to

Answer 1

We have $c=a\cos B+b\cos A$ (By projection rule)
and $a\sin B-b\sin A=0$. (By sine rule)
Thus, we can write $c=a(\cos B-i\sin B)+b(\cos A+i\sin A)$ where $i=\sqrt{-1}$
$\Rightarrow c^3={(a{e}^{-iB}+b{e}^{iA})}^3\Rightarrow c^3=a^3{e}^{-i3B}+3a^{2}b{e}^{-i2B}{e}^{iA}+3a{b}^2{e}^{-iB}{e}^{i2A}+b^3{e}^{i3A}\Rightarrow c^3=l$
(Consider real part only as in LHS, we have real value only)
$\therefore \frac{l}{c^3}=1$