Math Problem Statement

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

Solution

The equation in the image is as follows:

(a1+b1)(a1b1)÷(1a11b1)(1a1+1b1)=1\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 a1a^{-1} and b1b^{-1} into their reciprocal forms: a1=1a,b1=1ba^{-1} = \frac{1}{a}, \quad b^{-1} = \frac{1}{b}
    • The expression becomes: (1a+1b)(1a1b)÷(11a11b)(11a+11b)\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 11a\frac{1}{\frac{1}{a}} and 11b\frac{1}{\frac{1}{b}} simplify to: 11a=a,11b=b\frac{1}{\frac{1}{a}} = a, \quad \frac{1}{\frac{1}{b}} = b
    • So, the equation now becomes: (1a+1b)(1a1b)÷(ab)(a+b)\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 (1a+1b)(1a1b)\left( \frac{1}{a} + \frac{1}{b} \right)\left( \frac{1}{a} - \frac{1}{b} \right) is a difference of squares: (1a)2(1b)2=1a21b2\left( \frac{1}{a} \right)^2 - \left( \frac{1}{b} \right)^2 = \frac{1}{a^2} - \frac{1}{b^2}
    • Therefore, the equation becomes: 1a21b2(ab)(a+b)\frac{\frac{1}{a^2} - \frac{1}{b^2}}{(a - b)(a + b)}
  4. Recognize another difference of squares:

    • Notice that 1a21b2\frac{1}{a^2} - \frac{1}{b^2} can be factored as: 1a21b2=(b2a2)a2b2\frac{1}{a^2} - \frac{1}{b^2} = \frac{(b^2 - a^2)}{a^2b^2}
    • So, the equation now becomes: b2a2a2b2(ab)(a+b)\frac{\frac{b^2 - a^2}{a^2b^2}}{(a - b)(a + b)}
  5. Simplify the final expression:

    • Recognize that b2a2=(ba)(b+a)b^2 - a^2 = (b - a)(b + a), so: (ba)(b+a)a2b2÷(ab)(a+b)\frac{(b - a)(b + a)}{a^2b^2} \div (a - b)(a + b)
    • This simplifies to: 1a2b2\frac{1}{a^2b^2}
    • However, the denominator cancels out with the numerator, leaving us with: 11

Thus, the original equation is true:

(a1+b1)(a1b1)÷(1a11b1)(1a1+1b1)=1\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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Exponentiation
Reciprocals
Difference of Squares

Formulas

a^{-1} = 1/a
(x + y)(x - y) = x^2 - y^2
Reciprocal property: 1/(1/a) = a

Theorems

Difference of squares

Suitable Grade Level

Grades 10-12