I would first examine the roots of the first derivative for turning points:

\(\displaystyle f'(x)=4x^3-12x+8=4\left(x^3-3x+2\right)=0\)

We can see that \(x=1\) is a root of the cubic factor, and so let's use synthetic division to fully factor:

\(\displaystyle \begin{array}{c|rr}& 1 & 0 & -3 & 2 \\ 1 & & 1 & 1 & -2 \\ \hline & 1 & 1 & -2 & 0 \end{array}\)

And so:

\(\displaystyle f'(x)=4(x-1)(x^2+x-2)=4(x-1)^2(x+2)\)

And so our critical values are:

\(\displaystyle x\in\{-2,1\}\)

Because the root of \(f'(x)\) defined by \(x=1\) is of odd multiplicity, this is not a turning point, since the sign of the derivative does not change across it, and so it cannot be a local extremum. And so to determine the nature of the sole extremum, we may observe that:

\(\displaystyle f''(x)=4\left(3x^2-3\right)=12(x+1)(x-1)\)

And then:

\(\displaystyle f''(-2)>0\)

Thus, by the second derivative test, the extremum is a minimum. And since:

\(\displaystyle \lim_{x\to\pm\infty}f(x)=\infty\)

We know this minimum is a global minimum, hence:

\(\displaystyle f_{\min}=f(-2)=-27\)