To solve the equation \(x^3 - 6x^2 + 12x - 8 = 27\), start by setting the equation to zero:

\[
x^3 - 6x^2 + 12x - 8 - 27 = 0
\]

This simplifies to:

\[
x^3 - 6x^2 + 12x - 35 = 0
\]

To solve this cubic equation, we can use either factoring (if possible), synthetic division, or numerical methods if an exact solution is not easily found. First, check if there are any rational roots using the Rational Root Theorem. The potential rational roots are the factors of the constant term (-35) divided by the factors of the leading coefficient (1), which means checking the values \(\pm 1, \pm 5, \pm 7, \pm 35\).

Calculate \(f(x)\) for each value:

1. \(f(1) = 1^3 - 6 \cdot 1^2 + 12 \cdot 1 - 35 = 1 - 6 + 12 - 35 = -28\)
2. \(f(-1) = (-1)^3 - 6 \cdot (-1)^2 + 12 \cdot (-1) - 35 = -1 - 6 - 12 - 35 = -54\)
3. \(f(5) = 5^3 - 6 \cdot 5^2 + 12 \cdot 5 - 35 = 125 - 150 + 60 - 35 = 0\)

Since \(f(5) = 0\), 5 is a root of the equation.

Now, use synthetic division to divide the polynomial by \(x - 5\):

\[
\begin{array}{r|rrrr}
5 & 1 & -6 & 12 & -35 \\
& & 5 & -5 & 35 \\
\hline
& 1 & -1 & 7 & 0 \\
\end{array}
\]

The quotient is \(x^2 - x + 7\).

Now, solve for roots of the quadratic polynomial \(x^2 - x + 7 = 0\) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \(a = 1\), \(b = -1\), \(c = 7\):

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1}
\]
\[
x = \frac{1 \pm \sqrt{1 - 28}}{2}
\]
\[
x = \frac{1 \pm \sqrt{-27}}{2}
\]
\[
x = \frac{1 \pm \sqrt{27}i}{2}
\]
\[
x = \frac{1 \pm 3\sqrt{3}i}{2}
\]

Thus, the roots of the cubic equation are:

\[ x = 5, \quad x = \frac{1 + 3\sqrt{3}i}{2}, \quad x = \frac{1 - 3\sqrt{3}i}{2} \]