Answer
$\dfrac{-2+i\sqrt{2}}{6}$
Work Step by Step
RECALL:
(1) $(a-b)^2 = a^2-2ab+b^2$
(2) $\sqrt{-1}=i$
(3) $i^2=-1$
(4) For any real number $a \gt 0$, $\sqrt{-a} = i\sqrt{a}$.
Use rule (4) above to obtain:
$=\dfrac{-8+i\sqrt{32}}{24}
\\=\dfrac{-8+i\sqrt{16(2)}}{24}
\\=\dfrac{-8+i\sqrt{4^2(2)}}{24}
\\=\dfrac{-8+i\cdot 4\sqrt{2}}{24}
\\=\dfrac{-8+4i\sqrt{2}}{24}$
Factor out 4 in the numerator then cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{4(-2+i\sqrt{2})}{24}
\\=\dfrac{\cancel{4}(-2+i\sqrt{2})}{\cancel{24}6}
\\=\dfrac{-2+i\sqrt{2}}{6}$
