- Identify the expression as a binomial square using the identity \((a - b)^2 = a^2 - 2ab + b^2\).
- Substitute \(a = 3\) and \(b = \frac{3}{x}\) into the identity.
- Calculate each individual term: \(3^2 = 9\), \(2 \cdot 3 \cdot \frac{3}{x} = \frac{18}{x}\), and \(\left(\frac{3}{x}\right)^2 = \frac{9}{x^2}\).
- Combine the terms to obtain the final result: \(\boxed{9 - \frac{18}{x} + \frac{9}{x^2}}\).
Explanation
- Identifying the Structure and Components
To expand and simplify the expression \(\left(3 - \frac{3}{x}\right)^2\), we recognize it as the square of a binomial. A binomial of the form \((a - b)^2\) follows the algebraic identity:
\[(a - b)^2 = a^2 - 2ab + b^2\]
In this specific case, we identify \(a = 3\) and \(b = \frac{3}{x}\). The expression is well-defined as long as \(x
eq 0\), which is a constraint we must keep in mind throughout the simplification process.
- Expanding the Expression
Now, we substitute \(a = 3\) and \(b = \frac{3}{x}\) into the identity \(a^2 - 2ab + b^2\):
\[\left(3 - \frac{3}{x}\right)^2 = 3^2 - 2 \cdot 3 \cdot \left(\frac{3}{x}\right) + \left(\frac{3}{x}\right)^2\]
Next, we perform the arithmetic for each term:
- \(a^2 = 3^2 = 9\)
- \(2ab = 2 \cdot 3 \cdot \frac{3}{x} = \frac{18}{x}\)
- \(b^2 = \left(\frac{3}{x}\right)^2 = \frac{9}{x^2}\)
- Combining and Simplifying Terms
Substituting the evaluated terms back into the identity, we get:
\[\left(3 - \frac{3}{x}\right)^2 = 9 - \frac{18}{x} + \frac{9}{x^2}\]
This expression is now fully expanded. We can verify that no further simplification is needed as the terms \(9\), \(-\frac{18}{x}\), and \(\frac{9}{x^2}\) are in their simplest forms. Therefore, the expanded form is \(\boxed{9 - \frac{18}{x} + \frac{9}{x^2}}\).
Examples
Expanding algebraic expressions is a foundational skill used in physics and engineering. For example, when calculating the power dissipated in a circuit where the voltage and resistance are related by a variable, you might encounter expressions involving squares of rational functions. By expanding these expressions, you can more easily analyze how the power changes with respect to the variable component. This simplifies the process of finding maximum or minimum power points in practical engineering designs.