1952 AHSME Problems/Problem 26

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.)

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
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1952 AHSME Problems/Problem 26

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