Difference between revisions of "1952 AHSME Problems/Problem 26"
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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 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
