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Difficulty: MediumProperties of Exponents in Algebraic Expressions

Which of the following expressions is equivalent to a−2−b−2a−1−b−1\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}} for all non-zero real numbers aa and bb where a≠ba \neq b?

  1. A
    a2−b2a^2 - b^2
  2. B
    1a+b\frac{1}{a+b}
  3. a+bab\frac{a+b}{ab}Answer
  4. D
    1a−b\frac{1}{a-b}
  5. E
    a+ba+b

Answer

The correct answer is a+bab\frac{a+b}{ab} because simplifying the original expression using the difference of squares or by rewriting negative exponents as fractions leads directly to this value.
By applying the difference of squares factorization to the numerator, the expression simplifies to a−1+b−1a^{-1} + b^{-1}. Converting these to standard fractions and finding a common denominator yields a+bab\frac{a+b}{ab}.

Step-by-Step Solution

1
Factor the numerator as a difference of squares.
a−2−b−2=(a−1)2−(b−1)2=(a−1−b−1)(a−1+b−1)a^{-2} - b^{-2} = (a^{-1})^2 - (b^{-1})^2 = (a^{-1} - b^{-1})(a^{-1} + b^{-1})
Since the terms are squared with negative exponents, we can use the difference of squares identity x2−y2=(x−y)(x+y)x^2 - y^2 = (x - y)(x + y) where x=a−1x = a^{-1} and y=b−1y = b^{-1}.
2
Substitute the factored numerator back into the expression and cancel the common factor.
(a−1−b−1)(a−1+b−1)a−1−b−1=a−1+b−1\frac{(a^{-1} - b^{-1})(a^{-1} + b^{-1})}{a^{-1} - b^{-1}} = a^{-1} + b^{-1}
Since a≠ba \neq b, we know a−1−b−1≠0a^{-1} - b^{-1} \neq 0, allowing us to divide out the common factor in the numerator and denominator.
3
Rewrite the expression using positive exponents and find a common denominator.
a−1+b−1=1a+1b=bab+aab=a+baba^{-1} + b^{-1} = \frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab} = \frac{a+b}{ab}
A negative exponent x−nx^{-n} represents the reciprocal 1xn\frac{1}{x^n}. Finding the common denominator abab allows us to combine the two fractions into a single expression.

Key Concept

Properties of negative exponents and difference of squares factorization

Alternative Method

Convert all terms to fractions with positive exponents first: write the numerator as 1a2−1b2=b2−a2a2b2\frac{1}{a^2} - \frac{1}{b^2} = \frac{b^2 - a^2}{a^2b^2} and the denominator as 1a−1b=b−aab\frac{1}{a} - \frac{1}{b} = \frac{b-a}{ab}. Then, divide the two complex fractions by multiplying the numerator by the reciprocal of the denominator: (b−a)(b+a)a2b2×abb−a\frac{(b-a)(b+a)}{a^2b^2} \times \frac{ab}{b-a}. After canceling the common factors (b−a)(b-a) and abab, the expression simplifies directly to a+bab\frac{a+b}{ab}.
Estimated Time:1m 30s
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Topics

Properties of Exponents in Algebraic Expressions

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