Difference between revisions of "1955 AHSME Problems/Problem 13"

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==Solution==
 
==Solution==
By the difference of squares property, <math>a^{-4} - b^{-4}</math> is equivalent to <math>(a^{-2} + b^{-2})(a^{-2} - b^{-2})</math>. This means the fraction is now equal to <math>\frac{(a^{-2} + b^{-2})(a^{-2} - b^{-2})}{a^{-2} - b^{-2}}</math>, which simplifies to <math>\textbf{(C)}\ a^{-2}+b^{-2}</math>
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By the difference of squares property, <math>a^{-4} - b^{-4}</math> is equivalent to <math>(a^{-2} + b^{-2})(a^{-2} - b^{-2})</math>. This means the fraction is now equal to <math>\frac{(a^{-2} + b^{-2})(a^{-2} - b^{-2})}{a^{-2} - b^{-2}}</math>, which simplifies to <math>\textbf{(C)}\ a^{-2}+b^{-2}</math>  
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== See Also ==
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{{AHSME box|year=1955|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 13:23, 1 August 2020

Problem 13

The fraction $\frac{a^{-4}-b^{-4}}{a^{-2}-b^{-2}}$ is equal to:

$\textbf{(A)}\ a^{-6}-b^{-6}\qquad\textbf{(B)}\ a^{-2}-b^{-2}\qquad\textbf{(C)}\ a^{-2}+b^{-2}\\ \textbf{(D)}\ a^2+b^2\qquad\textbf{(E)}\ a^2-b^2$

Solution

By the difference of squares property, $a^{-4} - b^{-4}$ is equivalent to $(a^{-2} + b^{-2})(a^{-2} - b^{-2})$. This means the fraction is now equal to $\frac{(a^{-2} + b^{-2})(a^{-2} - b^{-2})}{a^{-2} - b^{-2}}$, which simplifies to $\textbf{(C)}\ a^{-2}+b^{-2}$

See Also

1955 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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


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