Difference between revisions of "1951 AHSME Problems/Problem 40"
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==Problem== | ==Problem== | ||
+ | |||
+ | <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: | ||
+ | |||
+ | <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> | ||
+ | <math> \textbf{(E)}\ [(x^{3}-1)^{2}]^{2} </math> | ||
+ | |||
==Solution== | ==Solution== | ||
− | + | First, note that we can pull the exponents out of every factor, since they are all squared. This results in | |
+ | <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> | ||
+ | Now, multiplying the numerators together gives | ||
+ | <math>\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1}{x^3-1}\right)^{4}</math>, | ||
+ | which simplifies to <math>\boxed{1\textbf{ (C)}}</math>. | ||
+ | |||
== 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 13:36, 19 April 2014
Problem
equals:
Solution
First, note that we can pull the exponents out of every factor, since they are all squared. This results in
Now, multiplying the numerators together gives
,
which simplifies to
.
See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
Followed by Problem 41 | |
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