1952 AHSME Problems/Problem 26

Revision as of 17:01, 18 April 2014 by Hukilau17 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

If $\left(r+\frac1r\right)^2=3$, then $r^3+\frac1{r^3}$ equals

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6$

Solution

We know $r+\frac1r=\sqrt3$. Cubing this gives $r^3+3r+\frac3r+\frac1{r^3}=3\sqrt3$. But $3r+\frac3r=3\left(r+\frac1r\right)=3\sqrt3$, so subtracting this from the first equation gives $r^3+\frac1{r^3}=\boxed{0\textbf{ (C)}}$. (Actually, $r+\frac1r$ could have been equal to $-\sqrt3$ instead of $\sqrt3$, but this would have led to the same answer. Also, this answer implies that $r^6=-1$, which means that $r$ is a complex number.)

See also

1952 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png