For all non-zero real numbers a and b such that a \neq b and a \neq -b, which of the follo… — GRE General Test

For all non-zero real numbers

a

and

b

such that

a \neq b

and

a \neq -b

, which of the following expressions is equivalent to

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}} \div \frac{a^2 + ab + b^2}{a^3 - b^3}?

\frac{a^2 - b^2}{ab}

Simplifying

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}}

yields

\frac{a + b}{ab}

after canceling

(b - a)

from both numerator and denominator. Simplifying

\frac{a^2 + ab + b^2}{a^3 - b^3}

yields

\frac{1}{a - b}

using the difference of cubes identity. Dividing

\frac{a + b}{ab}

\frac{1}{a - b}

gives

\frac{a + b}{ab} \cdot (a - b) = \frac{a^2 - b^2}{ab}

Simplify the first rational expression

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}}

\frac{\frac{1}{a^2} - \frac{1}{b^2}}{\frac{1}{a} - \frac{1}{b}} = \frac{\frac{b^2 - a^2}{a^2 b^2}}{\frac{b - a}{ab}} = \frac{(b - a)(b + a)}{a^2 b^2} \cdot \frac{ab}{b - a} = \frac{a + b}{ab}

Convert negative exponents to fractions, find common denominators, and cancel the common non-zero factor

(b - a)

Simplify the second rational expression

\frac{a^2 + ab + b^2}{a^3 - b^3}

\frac{a^2 + ab + b^2}{(a - b)(a^2 + ab + b^2)} = \frac{1}{a - b}

Factor the denominator using the difference of cubes identity

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Divide the simplified first expression by the simplified second expression

\frac{a + b}{ab} \div \frac{1}{a - b} = \frac{a + b}{ab} \cdot (a - b) = \frac{(a + b)(a - b)}{ab} = \frac{a^2 - b^2}{ab}

Multiply by the reciprocal of the second expression and apply the difference of squares identity.

Simplifying complex fractions and factoring using difference of squares and difference of cubes identities.

Question