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

(Added page)
 
Line 11: Line 11:
 
==Solution==
 
==Solution==
  
{{solution}}
+
<math> \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}}</math>, <math>\boxed{\text{B}}</math>.
  
 
==See also==
 
==See also==
  
 
{{AHSME box|year=1958|num-b=2|num-a=4}}
 
{{AHSME box|year=1958|num-b=2|num-a=4}}

Revision as of 12:18, 3 June 2011

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

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

See also

1958 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions