$Yes, you can use the cosine rule if you want. Also, for the solution's method, if $a=\mathrm{cis} \left( \frac{\pi}{3}\right)$, note that then $a^3 = \mathrm{cis} \left(\pi\right)= -1$ and $a\neq -1$, so $a^3 + 1 = 0 \Rightarrow (a+1)(a^2 - a + 1) = 0 \Rightarrow a^2 - a + 1 = 0$, as $a\neq...