To solve the equation \((x + \frac{1}{x})^2 = 3\), we start by expanding the left side:
\[ (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2} \]
Setting this equal to 3 gives us:
\[ x^2 + 2 + \frac{1}{x^2} = 3 \]
Subtracting 2 from both sides leads to:
\[ x^2 + \frac{1}{x^2} = 1 \]
Next, we want to find the value of \(x^3 + \frac{1}{x^3}\). We can use the identity:
\[ x^3 + \frac{1}{x^3} = (x + \frac{1}{x}) (x^2 - 1 + \frac{1}{x^2}) \]
First, we need to find \(x + \frac{1}{x}\). To get \(x + \frac{1}{x}\), we take the square root of both sides of our earlier result:
\[ x + \frac{1}{x} = \sqrt{3} \]
(We take the positive root since \(x\) can be any real number; taking \(-\sqrt{3}\) gives us a similar result due to the even powers in \(x^2\) and \(\frac{1}{x^2}\).)
Next, we need to 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 substitute back into our expression for \(x^3 + \frac{1}{x^3}\):
\[ x^3 + \frac{1}{x^3} = (x + \frac{1}{x}) (x^2 - 1 + \frac{1}{x^2}) = \sqrt{3} \cdot 0 = 0 \]
Thus, the value of \(x^3 + \frac{1}{x^3}\) is:
\[ \boxed{0} \]