1952 AHSME Problems/Problem 26: Difference between revisions
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== Problem== | |||
If <math>\left(r+\frac1r\right)^2=3</math>, then <math>r^3+\frac1{r^3}</math> equals | |||
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6</math> | |||
==Solution== | |||
We know <math>r+\frac1r=\sqrt3</math>. Cubing this gives <math>r^3+3r+\frac3r+\frac1{r^3}=3\sqrt3</math>. But <math>3r+\frac3r=3\left(r+\frac1r\right)=3\sqrt3</math>, so subtracting this from the first equation gives | |||
<math>r^3+\frac1{r^3}=\boxed{0\textbf{ (C)}}</math>. | |||
(Actually, <math>r+\frac1r</math> could have been equal to <math>-\sqrt3</math> instead of <math>\sqrt3</math>, but this would have led to the same answer. Also, this answer implies that <math>r^6=-1</math>, which means that <math>r</math> is a complex number.) | |||
==See also== | |||
{{AHSME 50p box|year=1952|num-b=25|num-a=27}} | |||
{{MAA Notice}} | |||
Latest revision as of 17:01, 18 April 2014
Problem
If 
, then 
equals
Solution
We know 
. Cubing this gives 
. But 
, so subtracting this from the first equation gives
.
(Actually, 
could have been equal to 
instead of 
, but this would have led to the same answer. Also, this answer implies that 
, which means that 
is a complex number.)
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
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| All AHSME Problems and Solutions | ||
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