Is it complex?

Observe that on expanding the above summation we get

$\binom{3n}{1} -3\times \binom{3n}{3} + 3^2\times \binom{3n}{5} - 3^3 \times \binom{3n}{7} ....$

with the -ve sign occurring when the power of $3$ is odd, this gives us a hint to introduce $i$ in the question. Also the exponents of $3$ increase by $1$ after $2$ terms so we can conclude to use $\sqrt{3}$ in the expansion

Thus, after whacking brains using expansion of $(1 + \sqrt{3}i)^{3n}$ we get:

$(1 + \sqrt{3}i)^{3n} = \binom{3n}{0} + \binom{3n}{1}\times i\sqrt3 - \binom{3n}{2}\times 3 - \binom{3n}{3}\times i\times (3\sqrt3) + \binom{3n}{4}\times 9 .....$

Now dividing the expression by $i\sqrt3$ we get:

$\displaystyle \large \dfrac{(1 + \sqrt{3}i)^{3n}}{ i\sqrt3}$ $= \dfrac{\binom{3n}{0}}{i\sqrt3} + \binom{3n}{1} - \dfrac{\binom{3n}{2}}{i\sqrt3}\times 3 - \binom{3n}{3}\times 3 + \dfrac{\binom{3n}{4}\times 9}{i\sqrt3} .....$

This makes sure that we are going in a right way as our required terms are matching. Now to eliminate the remaining terms we use the expansion:

$(1 - \sqrt{3}i)^{3n} = \binom{3n}{0} - \binom{3n}{1}\times i\sqrt3 - \binom{3n}{2}\times 3 + \binom{3n}{3}\times i\times (3\sqrt3) + \binom{3n}{4}\times 9 .....$

Similarly dividing by $i\sqrt3$ we get:

$\displaystyle \large \dfrac{(1 - \sqrt{3}i)^{3n}}{ i\sqrt3}$ $= \dfrac{\binom{3n}{0}}{i\sqrt3} - \binom{3n}{1} - \dfrac{\binom{3n}{2}}{i\sqrt3}\times 3 + \binom{3n}{3}\times 3 + \dfrac{\binom{3n}{4}\times 9}{i\sqrt3} .....$

So the terms we need are of the opposite sign as of the above series so subtracting the series:

$\large \dfrac{(1 + \sqrt{3}i)^{3n}}{ i\sqrt3} - \dfrac{(1 - \sqrt{3}i)^{3n}}{ i\sqrt3} = 2\times ( \binom{3n}{1} -3\times \binom{3n}{3} + 3^2\times \binom{3n}{5} - 3^3 \times \binom{3n}{7} ....)$

$\therefore$ We are getting our LHS correct. Now evaluating RHS Multiplying and dividing by $2^{3n}$

$\large (\dfrac{(\dfrac{1}{2} + \dfrac{\sqrt{3}i}{2})^{3n}}{ i\sqrt3}) - \dfrac{(\dfrac{1}{2} - \dfrac{\sqrt{3}i)^{3n}}{2}}{ i\sqrt3})\times 2^{3n}$

$\rightarrow \large \dfrac{2^{3n}}{i\sqrt3}\times ( (e^{\frac{i\pi}{3}})^{3n} - (e^{\frac{-i\pi}{3}})^{3n}$

$\rightarrow \large \dfrac{2^{3n}}{i\sqrt3}\times ( e^{i3\pi n} - e^{-i3\pi n} )$

$\rightarrow \large \dfrac{2^{3n}}{i\sqrt3}\times ( (-1) - (-1)) = \boxed{0}$