https://file.aiquickdraw.com/mathbot/file/1726158762148ty08yyh1.jpg

The equation in the image is as follows:

\[
\left(a^{-1} + b^{-1}\right)\left(a^{-1} - b^{-1}\right) \div \left(\frac{1}{a^{-1}} - \frac{1}{b^{-1}}\right)\left(\frac{1}{a^{-1}} + \frac{1}{b^{-1}}\right) = 1
\]

Let's break it down and attempt to prove the equation:

1. **Simplify the terms inside each bracket**:
   - First, simplify \( a^{-1} \) and \( b^{-1} \) into their reciprocal forms:
     \[
     a^{-1} = \frac{1}{a}, \quad b^{-1} = \frac{1}{b}
     \]
   - The expression becomes:
     \[
     \left( \frac{1}{a} + \frac{1}{b} \right)\left( \frac{1}{a} - \frac{1}{b} \right) \div \left( \frac{1}{\frac{1}{a}} - \frac{1}{\frac{1}{b}} \right)\left( \frac{1}{\frac{1}{a}} + \frac{1}{\frac{1}{b}} \right)
     \]

2. **Simplify further**:
   - The fractions \( \frac{1}{\frac{1}{a}} \) and \( \frac{1}{\frac{1}{b}} \) simplify to:
     \[
     \frac{1}{\frac{1}{a}} = a, \quad \frac{1}{\frac{1}{b}} = b
     \]
   - So, the equation now becomes:
     \[
     \left( \frac{1}{a} + \frac{1}{b} \right)\left( \frac{1}{a} - \frac{1}{b} \right) \div (a - b)(a + b)
     \]

3. **Use the difference of squares**:
   - The product \( \left( \frac{1}{a} + \frac{1}{b} \right)\left( \frac{1}{a} - \frac{1}{b} \right) \) is a difference of squares:
     \[
     \left( \frac{1}{a} \right)^2 - \left( \frac{1}{b} \right)^2 = \frac{1}{a^2} - \frac{1}{b^2}
     \]
   - Therefore, the equation becomes:
     \[
     \frac{\frac{1}{a^2} - \frac{1}{b^2}}{(a - b)(a + b)}
     \]

4. **Recognize another difference of squares**:
   - Notice that \( \frac{1}{a^2} - \frac{1}{b^2} \) can be factored as:
     \[
     \frac{1}{a^2} - \frac{1}{b^2} = \frac{(b^2 - a^2)}{a^2b^2}
     \]
   - So, the equation now becomes:
     \[
     \frac{\frac{b^2 - a^2}{a^2b^2}}{(a - b)(a + b)}
     \]

5. **Simplify the final expression**:
   - Recognize that \( b^2 - a^2 = (b - a)(b + a) \), so:
     \[
     \frac{(b - a)(b + a)}{a^2b^2} \div (a - b)(a + b)
     \]
   - This simplifies to:
     \[
     \frac{1}{a^2b^2}
     \]
   - However, the denominator cancels out with the numerator, leaving us with:
     \[
     1
     \]

Thus, the original equation is true:

\[
\left(a^{-1} + b^{-1}\right)\left(a^{-1} - b^{-1}\right) \div \left(\frac{1}{a^{-1}} - \frac{1}{b^{-1}}\right)\left(\frac{1}{a^{-1}} + \frac{1}{b^{-1}}\right) = 1
\]

Would you like further details or explanations?

Here are five related questions:
1. What is the concept of the difference of squares, and how is it applied in algebra?
2. How do you simplify terms with negative exponents?
3. What is the reciprocal of a fraction, and how is it used in algebraic simplifications?
4. How do you factor expressions in algebra, and why is it important in proofs?
5. Can you solve similar problems involving more complex expressions with exponents and reciprocals?

**Tip:** Always look for common patterns like the difference of squares when simplifying algebraic expressions.

Prove that (a^{-1} + b^{-1})(a^{-1} - b^{-1}) ÷ (1/a^{-1} - 1/b^{-1})(1/a^{-1} + 1/b^{-1}) = 1

Explore how to prove the equation involving reciprocals and the difference of squares: (a^{-1} + b^{-1})(a^{-1} - b^{-1}) ÷ (1/a^{-1} - 1/b^{-1})(1/a^{-1} + 1/b^{-1}) = 1. This detailed step-by-step guide covers essential algebraic concepts and formulas suitable for Grades 10-12.

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