Difference between revisions of "1958 AHSME Problems/Problem 3"

Latest revision as of 05:10, 3 October 2014

Problem

Of the following expressions the one equal to $\frac{a^{-1}b^{-1}}{a^{-3} - b^{-3}}$ is:

$\textbf{(A)}\ \frac{a^2b^2}{b^2 - a^2}\qquad \textbf{(B)}\ \frac{a^2b^2}{b^3 - a^3}\qquad \textbf{(C)}\ \frac{ab}{b^3 - a^3}\qquad \textbf{(D)}\ \frac{a^3 - b^3}{ab}\qquad \textbf{(E)}\ \frac{a^2b^2}{a - b}$

Solution

$\frac{a^{-1}b^{-1}}{a^{-3} - b^{-3}} = \frac{\frac{1}{ab}}{\frac{1}{a^{3}}-\frac{1}{b^{3}}} = \frac{\frac{1}{ab}}{\frac{1}{a^{3}}-\frac{1}{b^{3}}}\cdot\frac{a^{3}b^{3}}{a^{3}b^{3}} = \frac{a^{2}b^{2}}{b^{3}-a^{3}}$ , $\boxed{\text{B}}$ .

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 15: / Line 15: @@
 ==See also==
-{{AHSME box|year=1958|num-b=2|num-a=4}}
+{{AHSME 50p box|year=1958|num-b=2|num-a=4}}
+{{MAA Notice}}

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Difference between revisions of "1958 AHSME Problems/Problem 3"

Latest revision as of 05:10, 3 October 2014

Problem

Solution

See also