What is the Galois group of 27 x 8 − 72 x 4 − 16...

The roots are easily found to be
$$\begin{gathered} \pm \sqrt[4]{\frac{4(3+2\sqrt{3})}{9}}=\pm a\text{ (say)}, \\ \pm i\sqrt[4]{\frac{4(3+2\sqrt{3})}{9}}, \\ \pm (1\pm i)\sqrt[4]{\frac{2\sqrt{3}-3}{9}}=\pm (1\pm i)b\text{ (say)} \end{gathered}$$
with a,b being positive real numbers.
The given polynomial is irreducible over $\mathbb{Q}$ and hence $[\mathbb{Q}(a):\mathbb{Q}]=8$. The splitting field L of the polynomial contains $a,ai$ and hence $i\in L$. But $i\notin \mathbb{Q}(a)$ and hence $[\mathbb{Q}(a,i):\mathbb{Q}]=16$
It can be proved with some effort that $L=\mathbb{Q}(a,i)$ is the splitting field of the polynomial and hence the Galois group is of order 16.
We can note that
$ab=\frac{\sqrt[4]{12}}{3}=\frac{c}{3}\text{ (say)}$
and next we show that $c\in L$. Since c is real we have in fact $c\in \mathbb{Q}(a)$. We can observe that
$c^2=2\sqrt{3}=\frac{9a^4-12}{4}\in \mathbb{Q}(a)$
and
$\begin{array}{rl}9a^4& =4c^2+12\\ & =4c^2+c^4\\ & =c^2(c^2+4)\\ & =c^2(4+2\sqrt{3})\\ & =c^2(1+\sqrt{3}{)}^2\end{array}$
This implies
$3a^2=c(1+\sqrt{3})$
ie
$c=\frac{6a^2}{2+c^2}\in \mathbb{Q}(a)$
It now follows that $b=c/(3a)\in \mathbb{Q}(a)\subset L$ and L is the desired splitting field.
The Galois group is $D_{16}$ (dihedral group of order 16) as explained in another answer.