A somewhat more systematic way to do this problem is to write the number you're taking powers of in polar form, as
$$ (-\sqrt{2} + i \sqrt{6}) = \sqrt{8} {-1 + i \sqrt{3} \over 2} = \sqrt{8} \left( {\cos {2\pi \over 3}} + i {\sin {2\pi \over 3}} \right) = \sqrt{8} e^{-2i \pi/3}$$.
Therefore
$$ (-\sqrt{2} + i \sqrt{6})^n = 8^{n/2} e^{-2i \pi/3 \times n}. $$
For this to be an integer, we must have that $8^{n/2}$ is an integer and that $2 \pi/3 \times n$ is a multiple of $\pi$ (therefore making the whole thing a real number). The first condition is satisfied if $n$ is even; the second is satisfied if $n$ is a multiple of 3. In particular $n = 6$ is the smallest $n$ that will work, and
$$ (-\sqrt{2} + i \sqrt{6})^6 = 8^3 e^{-2i\pi} = 8^3.$$