Problem when $x=\cos (a) +i\sin(a),\ y=\cos (b) +i\sin(b),\ z=\cos (c) +i\sin(c),\ x+y+z=0$ Problem: If $$x=\cos (a) +i\sin(a),\ y=\cos (b) +i\sin(b),\ z=\cos (c) +i\sin(c),\ x+y+z=0$$ then which of the following can be true:
1) $\cos 3a + \cos 3b + \cos 3c = 3 \cos (a+b+c)$
2) $1+\cos (a-b) + \cos (b-c) =0$
3) $\cos 2a + \cos 2b +\cos 2c =\sin 2a +\sin 2b +\sin 2c=0$
4) $\cos (a+b)+\cos(b+c)+\cos(c+a)=0$
Try: I tried taking $x=e^{ia},y=e^{ib},z=e^{ic}$ and then i tried expressing each option in euler form 
FOR EXAMPLE:
1) $-3/2 e^{-i a-i b-i c}-3/2 e^{i a+i b+i c}+1/2 e^{-3 i a}+1/2 e^{3 i a}+1/2 e^{-3 i b}+1/2 e^{3 i b}+1/2 e^{-3 i c}+1/2 e^{3 i c}$
2) $1/2 e^{i a-i b}+1/2 e^{i b-i a}+1/2 e^{i b-i c}+1/2 e^{i c-i b}+1$
3) $1/2 e^{-2 i a}+1/2 e^{2 i a}+1/2 e^{-2 i b}+1/2 e^{2 i b}+1/2 e^{-2 i c}+1/2 e^{2 i c}$
4) $1/2 e^{-i a-i b}+1/2 e^{i a+i b}+1/2 e^{-i a-i c}+1/2 e^{i a+i c}+1/2 e^{-i b-i c}+1/2 e^{i b+i c}$
Now after all this i'm stuck!!Please help!! How should i proceed?

HINT:
For $(1),$
Using If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.,
we have $x^3+y^3+z^3=3xyz$
Now $x^3=e^{i(3a)}=\cos3a+i\sin3a$
and $xy=e^{i(a+b)}=\cos(a+b)+i\sin(a+b),xyz=\cdots$
For $(3),(4)$
$x+y+z=0\implies\cos a+\cos b+\cos c=\sin a+\sin b+\sin c=0$
So, $x^{-1}+y^{-1}+z^{-1}=?$
Now $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=(x+y+z)^2-2xyz(x^{-1}+y^{-1}+z^{-1})=?$
