We are given the equation:

$a + \frac{1}{a} = 3 \frac{1}{3} = \frac{10}{3}$

We need to find the value of:

$a^3 - \frac{1}{a^3}$

Step 1: Let $a + \frac{1}{a} = x$

Given that $x = \frac{10}{3}$, we now need to express $a^3 - \frac{1}{a^3}$ in terms of $x$.

Step 2: Use the identity for cubes

The identity that relates $a^3 - \frac{1}{a^3}$ to $a + \frac{1}{a}$ is:

$a^3 - \frac{1}{a^3} = \left( a + \frac{1}{a} \right) \left( a^2 - 1 + \frac{1}{a^2} \right)$

First, we need to find the value of $a^2 + \frac{1}{a^2}$. We can use the square of $a + \frac{1}{a}$:

$\left( a + \frac{1}{a} \right)^2 = a^2 + 2 + \frac{1}{a^2}$

So,

$\left( \frac{10}{3} \right)^2 = a^2 + 2 + \frac{1}{a^2}$

$\frac{100}{9} = a^2 + 2 + \frac{1}{a^2}$

Subtract 2 from both sides:

$a^2 + \frac{1}{a^2} = \frac{100}{9} - 2 = \frac{100}{9} - \frac{18}{9} = \frac{82}{9}$

Step 3: Find $a^3 - \frac{1}{a^3}$

Now, use the identity:

$a^3 - \frac{1}{a^3} = \left( a + \frac{1}{a} \right) \left( a^2 - 1 + \frac{1}{a^2} \right)$

Simplifying this, we find:

$a^3 - \frac{1}{a^3} = \left( \frac{10}{3} \right) \left( \frac{82}{9} \right)$

Multiply these fractions:

$a^3 - \frac{1}{a^3} = \frac{10 \times 82}{3 \times 9} = \frac{820}{27}$

Thus, the value of $a^3 - \frac{1}{a^3}$ is:

$\boxed{\frac{820}{27}}$

Would you like more details, or do you have any questions?

Here are some related questions to explore:

1. How does this formula change if $a + \frac{1}{a}$ equals a different value?
2. Can you derive similar identities for higher powers of $a$?
3. What if $a + \frac{1}{a}$ were a complex number?
4. How can you use the same approach for negative exponents?
5. Can you derive a general formula for $a^n - \frac{1}{a^n}$?

Tip: When solving equations with fractions, always consider simplifying expressions using identities for sums, products, and powers.