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A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

As stated the expression above should be equal to \([\frac{x^2-9}{3}*\frac{8}{x+3}]^{-1} = [\frac{(x-3)(x+3)}{3}*\frac{8}{x+3}]^{-1} = [\frac{(x-3)}{3}*8]^{-1} = \frac{3}{8}*\frac{1}{x-3}\). Thus, the answer is uncertain since \(\frac{1}{x-3}\) can be 1 making the two expression equal but also greater or lower than 1. How can the answer be B?

The first quantity is raised to -1, then switch it in the reverse form.

\(\frac{3(x+3)}{8(x+3)(x-3)}\)

(x+3) cancel out

\(\frac{3}{8(x-3)}\)

Since x > 4, the denominator in Quantity A must be greater than 8; since the numerators in Quantity A and Quantity B are the same and the denominator in Quantity A is larger, Quantity B must be greater.