This problem was from the 2004 Luzerne County Math Contest:
If (r + 1/r)^2 = 3, what is r + (1/r^3) = ?
Any help would really be appreciated. Thanks in advance!
I don't know if there is a more elegant way of solving this but:
You can solve
$\displaystyle \displaystyle \left( r+\frac{1}{r}\right)^2=3$
which has complex roots but which can be represented in exponential form quite nicely.
Substitute into:
$\displaystyle \displaystyle r+\left(\frac{1}{r}\right)^3$
and simplify, and if what I have done has no algebraic errors you should end up with $\displaystyle r=\pm i$
(note: in a competition you will get no marks for the final answer but just from the method you used to find it)
CB
Hello, eric1299171!
Is there a typo?
If the second expression has $\displaystyle \,r^3$ twice,
. . there is a rather elegant solution.
And it looks more like a contest problem.
$\displaystyle \text{Given: }\:\left(r + \dfrac{1}{r}\right)^2 \,=\, 3.$
. . . . . . . . ↓
$\displaystyle \text{Find: }\:r^3 + \dfrac{1}{r^3}$
We have: . . . . . $\displaystyle r + \dfrac{1}{r} \;=\;\sqrt{3}$
Then: . . . . . $\displaystyle \left(r + \dfrac{1}{r}\right)^3 \;=\;3\sqrt{3}$
. . . . $\displaystyle r^3 + 3r + \dfrac{3}{r} + \dfrac{1}{r^3} \;=\;3\sqrt{3}$
. . $\displaystyle \displaystyle r^3 + 3\underbrace{\left(r + \frac{1}{r}\right)}_{\text{This is }\sqrt{3}} + \frac{1}{r^3} \;=\;3\sqrt{3}$
. . . . . . $\displaystyle \displaystyle r^3 + 3\sqrt{3} + \frac{1}{r^3} \;=\;3\sqrt{3}$
Therefore: . . . .$\displaystyle r^3 + \dfrac{1}{r^3} \;=\;0$