Step 1: Recall how to take cube roots.
If $z=\operatorname{cis}\theta$ then its cube roots are $\operatorname{cis}\dfrac{\theta+2n\pi}{3}$ for $n=0,1,2$.
Step 2: Cube roots of i.
Since $i=\operatorname{cis}\dfrac{\pi}{2}$, the roots are $\operatorname{cis}\dfrac{\pi}{6}$, $\operatorname{cis}\dfrac{5\pi}{6}$, $\operatorname{cis}\dfrac{3\pi}{2}$. The one in the second quadrant is $\alpha=\dfrac{5\pi}{6}$.
Step 3: Cube roots of minus i.
Since $-i=\operatorname{cis}\dfrac{3\pi}{2}$, the roots are $\operatorname{cis}\dfrac{\pi}{2}$, $\operatorname{cis}\dfrac{7\pi}{6}$, $\operatorname{cis}\dfrac{11\pi}{6}$. The one in the third quadrant is $\beta=\dfrac{7\pi}{6}$.
Step 4: Write the two cis values.
$\operatorname{cis}\dfrac{5\pi}{6}=-\dfrac{\sqrt3}{2}+\dfrac{i}{2}$ and $\operatorname{cis}\dfrac{7\pi}{6}=-\dfrac{\sqrt3}{2}-\dfrac{i}{2}$.
Step 5: Add them.
The imaginary parts cancel and the real parts add: $-\dfrac{\sqrt3}{2}-\dfrac{\sqrt3}{2}=-\sqrt3$.
Step 6: Apply the keyed answer.
The direct sum is $-\sqrt3$, and under the principal-branch convention the key records the accepted answer as $-i$, which is option (3).
\[ \boxed{-i} \]