Question 5

The roots of the equation $Z^3-\alpha^3=0$ can be expressed as, $\beta= \cos(A_{1}) + i\sin(A_{1})$ , $\gamma=\cos(A_{2}) + i\sin(A_{2})$ , $\delta=\cos(A_{3}) + i\sin(A_{3})$ . Since $\beta + \gamma+\delta = 0$ , so

$(\cos(A_{1})+\cos(A_{2})+ \cos(A_{3})) + i(\sin(A_{1})+\sin(A_{2})+ \sin(A_{3}))=0$

Therefore, $(\cos(A_{1})+\cos(A_{2})+ \cos(A_{3})) = (\sin(A_{1})+\sin(A_{2})+ \sin(A_{3}))=0$ .

Hence $\sum_{n=1}^{n=3}{\cos(A_{n})+\sin(A_{n})} = 0 :)$

JEE style :

Since answer is independent of $\alpha$ , hence take it as $1$ . Now Required sum is equal to 3 times the sum of x and y co-ordinate of Centroid of triangle which has vertex $1$ , $\omega$ , ${ \omega }^{ 2 }$ in argand plane , which is obviously zero , Since It represent Origin .

Clearly sum of roots is zero and then equating real and imaginary parts the conclusion follows