<s>[INST] Finding various things given Cos(a-b)+Cos(b-c)+Cos(c-a) = $\frac{-3}{2}$ I have given the task to find $\sin^6(a)+\sin^6(b)+\sin^6(c)$. It is also given $a , b$ and $c$ are real .
Further given down is only my attempt , its your choice you want to read it not . I just want a solution which could be completing my solution or give a better and simpler method . Any help like another attempt , links , hints etc are appreciated. Solution can include use of complex numbers or trigonometry. It would be appreciated if you don't use tools higher than high school level .

First thing I noticed is he above equation can be written as $[\sin(a)+\sin(b)+\sin(c)]^2\ +\ [\cos(a)+\cos(b)+\cos(c)]^2\ =\ 0$
Which implies $\cos(a)+\cos(b)+\cos(c)=0$ .....(1) and also $\sin(a)+\sin(b)+\sin(c)=0$ ....(2) .

$NA  =\cos(Na)+ i\sin(Na)$
$NB  =\cos(Nb)+ i\sin(Nb)$
$NC  =\cos(Nc)+ i\sin(Nc)$
Where $N$ is an arbitrary natural number and $i$ is the square root of $-1$.

From (1) and (2)
$A+B+C=0 .. (3)$
Which implies $\frac{1}{AB} + \frac{1}{BC} + \frac{1}{BC}=0$
Taking conjugate both sides
$AB+BC+CA=0$ ....(4)
So $A²+B²+C²=0$ .....(5) also as $(A+B+C)^2=0$
We can also by repeating same procedure $A^4+B^4+C^4=0$ ....(6)

Now $\cos²(a)+\cos²(b)+\cos²(c)$ = $\frac{1}{2}[3-\cos(2a)-\cos(2b)-\cos(2c)]$ = $\frac{3}{2}$
As by (5) we know $\cos(2a)+\cos(2b)+\cos(2c)=0$
Same way $\sin^2(a)+\sin^2(b)+\sin^2(c)= \frac{3}{2}$
Also $\cos^2(2a)+\cos^2(2b)+\cos^2(2c)=\frac{3}{2}$ as $\cos(4a) + \cos(4b) + \cos(4c) = 0 $ by (6)
As $[\cos(a) + \cos(b) + \cos(c)]^2 = 0 $, so $\cos(a)\cos(b)+ \cos(b)\cos(c)+\cos(c)\cos(a)= \frac{-3}{4} $
Same way $\cos(2a)\cos(2b)+\cos(2b)\cos(2c) +\cos(2c)\cos(2a)= \frac{-3}{4} $

$[\sin^2(a)-\sin^2(b)]^2+ [\sin^2(b)-\sin^2(c)]^2+ [\sin^2(c)-\sin^2(a)]^2$
= $\frac{1}{4} \left([\cos(2a)-\cos(2b)]^2+ [\cos(2b)-\cos(2c)]^2 +[\cos(2c)-\cos(2a)]^2\right)$
= $\frac{9}{8}$

$l³+m³+n³=(l+m+n)(l²+m²+n²-lm-mn-nm) + 3lmn $
$l=\sin²(a),\:m=\sin²(b)$ and $n=\sin²(c)$
I have already found $l+m+n$ and $l²+m²+n²-lm-mn-nm$ above. So I am stuck with finding $lmn$.
Please help me find $lmn$ or provide another easier solution as you can see this method is really lenghty and tedious . Please also make out any mistakes I have made .
 [/INST] From your notes:
$$\cos a+\cos b+\cos c=0 \Rightarrow \cos^2a+\cos^2b+2\cos a\cos b=\cos^2c\\
\sin a+\sin b+\sin c=0\Rightarrow \sin^2a+\sin^2b+2\sin a\sin b=\sin^2c\\
$$
Adding the two:
$$\cos(a-b)=-\frac12.$$
Similarly:
$$\cos(b-c)=-\frac12\\
\cos(c-a)=-\frac12$$
One solution is: $a=180^\circ,b=60^\circ,c=-60^\circ$.
Hence, the sum of sixth powers of sines is $0$.
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