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Difference between revisions of "1951 AHSME Problems/Problem 40"

(Created page with "== See Also == {{AHSME 50p box|year=1951|num-b=39|num-a=41}} Category:Introductory Algebra Problems {{MAA Notice}}")
 
 
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==Problem==
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<math> \left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2} </math> equals:
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<math> \textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2} </math>
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<math> \textbf{(E)}\ [(x^{3}-1)^{2}]^{2} </math>
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==Solution==
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First, note that we can pull the exponents out of every factor, since they are all squared. This results in
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<math>\left(\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\right)^{4}\cdot\left(\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\right)^{4}</math>
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Now, multiplying the numerators together gives
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<math>\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1}{x^3-1}\right)^{4}</math>,
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which simplifies to <math>\boxed{1\textbf{ (C)}}</math>.
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== See Also ==
 
== See Also ==
 
{{AHSME 50p box|year=1951|num-b=39|num-a=41}}  
 
{{AHSME 50p box|year=1951|num-b=39|num-a=41}}  

Latest revision as of 12:36, 19 April 2014

Problem

$\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:

$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$ $\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$

Solution

First, note that we can pull the exponents out of every factor, since they are all squared. This results in $\left(\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\right)^{4}\cdot\left(\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\right)^{4}$ Now, multiplying the numerators together gives $\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1}{x^3-1}\right)^{4}$, which simplifies to $\boxed{1\textbf{ (C)}}$.

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Problem 41
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All AHSME Problems and Solutions

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