To solve for \( x^3 + \frac{1}{x^3} \) given the equation
\[ \left( x + \frac{1}{x} \right)^2 = 3, \]
we start by expanding the left side:
\[ x^2 + 2 + \frac{1}{x^2} = 3. \]
Subtracting 2 from both sides gives us:
\[ x^2 + \frac{1}{x^2} = 1. \]
Next, we will use the identity for \( x^3 + \frac{1}{x^3} \):
\[ x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right) \left( x^2 - 1 + \frac{1}{x^2} \right). \]
We already have \( x^2 + \frac{1}{x^2} \), which is 1. From this, we can find \( x^2 - 1 + \frac{1}{x^2} \):
\[ x^2 - 1 + \frac{1}{x^2} = (x^2 + \frac{1}{x^2}) - 1 = 1 - 1 = 0. \]
Now we also need to find \( x + \frac{1}{x} \):
To find \( x + \frac{1}{x} \), we can take the square root of both sides of the original equation. Since we have:
\[ (x + \frac{1}{x})^2 = 3, \]
it implies:
\[ x + \frac{1}{x} = \sqrt{3} \quad \text{or} \quad x + \frac{1}{x} = -\sqrt{3}. \]
However, either case leads to the same calculation since \( x^3 + \frac{1}{x^3} \) depends only on \( x + \frac{1}{x} \).
Now substituting into the identity:
\[ x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right) \left( x^2 - 1 + \frac{1}{x^2} \right), \] with \( (x + \frac{1}{x}) = \sqrt{3} \) (or \(-\sqrt{3}\)) and \( x^2 - 1 + \frac{1}{x^2} = 0 \):
\[ x^3 + \frac{1}{x^3} = \sqrt{3} \cdot 0 = 0. \]
Thus, the value of \( x^3 + \frac{1}{x^3} \) is:
\[ \boxed{0}. \]