To solve the equation \( x^3 - 3x^2 - 3x - 1 = 0 \), we can use various methods such as factoring, synthetic division, or numerical methods if necessary. Let's start by trying to find rational roots using the Rational Root Theorem.

The possible rational roots of the polynomial are the factors of the constant term (-1) divided by the factors of the leading coefficient (1). Therefore, the possible rational roots are \( \pm 1 \).

### Testing Possible Roots
1. **Testing \( x = 1 \)**:

\[
1^3 - 3(1^2) - 3(1) - 1 = 1 - 3 - 3 - 1 = -6 \quad (\text{not a root})
\]

2. **Testing \( x = -1 \)**:

\[
(-1)^3 - 3(-1)^2 - 3(-1) - 1 = -1 - 3 + 3 - 1 = -2 \quad (\text{not a root})
\]

### Finding Roots Numerically or Graphically
Since we didn't find a rational root, we can try using numerical methods or graphing to approximate the roots.

### Cubic Formula
Alternatively, we can apply Cardano's method for solving cubic equations. The general form of a cubic equation is:

\[
x^3 + ax^2 + bx + c = 0
\]

In your case:

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

Let's rewrite it in a depressed form by substituting \( x = y + \frac{a}{3} \).

1. Set \( a = -3 \), so we substitute:

\[
x = y + 1
\]

2. Rewrite the equation in terms of \( y \):

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

Expanding this we get:

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

This simplifies to:

\[
y^3 - 3 = 0
\]

Thus, we have:

\[
y^3 = 3 \implies y = \sqrt[3]{3}
\]

### Back Substituting
Recall that \( x = y + 1 \).

So:

\[
x = \sqrt[3]{3} + 1
\]

### Conclusion
In addition to that root, cubic equations can have three roots. The other two can be found using complex numbers if necessary. The full set can be derived from:

\[
x = \sqrt[3]{3} + 1, \quad x = \sqrt[3]{3}\omega + 1, \quad x = \sqrt[3]{3}\omega^2 + 1
\]

where \( \omega = e^{2\pi i / 3} \) is a complex cube root of unity.

Calculating these yields the solution:

- One real root: \( x = \sqrt[3]{3} + 1 \)
- Two complex roots: \( x = \sqrt[3]{3}\omega + 1 \) and \( x = \sqrt[3]{3}\omega^2 + 1 \).

If you need numerical approximations or further exploration, please let me know!