To express \(\frac{1}{x^3 - 1}\) in partial fractions, we need to factor the denominator \(x^3 - 1\) and then decompose the fraction into simpler fractions. The denominator \(x^3 - 1\) can be factored as the difference of cubes:

\[x^3 - 1 = (x - 1)(x^2 + x + 1)\]

Now, we can decompose the fraction using partial fractions:

\[\frac{1}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}\]

To solve for the constants \(A\), \(B\), and \(C\), we'll clear the denominators by multiplying both sides of the equation by \(x^3 - 1\):

\[1 = A(x^2 + x + 1) + (Bx + C)(x - 1)\]

Expand both sides:

\[1 = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C\]

Now, group the terms with the same powers of \(x\):

\[1 = (A + B)x^2 + (A - B + C)x + (A - C)\]

Equating coefficients on both sides, we get the following system of equations:

\[A + B = 0\]

\[A - B + C = 0\]

\[A - C = 1\]

Solving this system, we find:

\(A = \frac{1}{3}\)

\(B = -\frac{1}{3}\)

\(C = -\frac{2}{3}\)

Now, substitute the values of \(A\), \(B\), and \(C\) back into the partial fraction decomposition:

\[\frac{1}{x^3 - 1} = \frac{1}{3(x - 1)} - \frac{1}{3}\frac{x + 2}{x^2 + x + 1}\]

Simplify the second term on the right side:

\[\frac{1}{3}\left(\frac{1}{x - 1} - \frac{x + 2}{x^2 + x + 1}\right)\]