Algebraic Solution for Fraction Equation

We are given that $p$ and $q$ are positive integers and the equation:

$ \frac{p}{q} + \frac{q}{p} = 3 $

We need to find the value of:

$ \frac{p^2}{q^2} + \frac{q^2}{p^2} $

Step-by-Step Calculation

1. Let $x = \frac{p}{q}$. The given equation can be rewritten in terms of $x$ as:

   $ x + \frac{1}{x} = 3 $

2. The expression we need to find is $\frac{p^2}{q^2} + \frac{q^2}{p^2}$, which is equivalent to $x^2 + \frac{1}{x^2}$.

3. Consider the square of the term $(x + \frac{1}{x})$:

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

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

4. Rearrange the formula to solve for $x^2 + \frac{1}{x^2}$:

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

5. Substitute the given value $\left(x + \frac{1}{x} = 3\right)$ into the rearranged formula:

   $ x^2 + \frac{1}{x^2} = (3)^2 - 2 $

   $ x^2 + \frac{1}{x^2} = 9 - 2 $

   $ x^2 + \frac{1}{x^2} = 7 $

6. Since $x = \frac{p}{q}$, we have $x^2 = \frac{p^2}{q^2}$ and $\frac{1}{x^2} = \frac{q^2}{p^2}$. Therefore:

   $ \frac{p^2}{q^2} + \frac{q^2}{p^2} = 7 $

Conclusion

The value of $\frac{p^2}{q^2} + \frac{q^2}{p^2}$ is 7.