To solve the equation \(x^3 - x(x+1) + 1 = 0\), let's first expand and simplify it.

1. Expand the term \(x(x+1)\):

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

2. Substitute back into the original equation:

\[
x^3 - (x^2 + x) + 1 = 0
\]

3. Simplify the equation:

\[
x^3 - x^2 - x + 1 = 0
\]

Now, we have the simplified equation \(x^3 - x^2 - x + 1 = 0\). To solve a cubic equation like this, we can look for rational roots using the Rational Root Theorem or try factoring by grouping.

### Factoring by Grouping

1. Let's group terms to see if we can factor the equation:

\[
(x^3 - x^2) - (x - 1) = 0
\]

2. Factor each group:

\[
x^2(x - 1) - 1(x - 1) = 0
\]

3. Notice we have a common factor of \(x - 1\):

\[
(x^2 - 1)(x - 1) = 0
\]

4. Further factor \(x^2 - 1\) as a difference of squares:

\[
(x - 1)(x + 1)(x - 1) = 0
\]

5. Simplify the expression:

\[
(x - 1)^2 (x + 1) = 0
\]

From this factorization, the solutions to the equation are:

- \(x - 1 = 0 \Rightarrow x = 1\) (with multiplicity 2)
- \(x + 1 = 0 \Rightarrow x = -1\)

Therefore, the solutions to the equation \(x^3 - x(x+1) + 1 = 0\) are \(x = 1\) (with multiplicity 2) and \(x = -1\).