# Thread: Math Contest Question 4

1. ## Math Contest Question 4

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!

2. Originally Posted by eric1299171
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

$\displaystyle \left( r+\frac{1}{r}\right)^2=3$

which has complex roots but which can be represented in exponential form quite nicely.

Substitute into:

$\displaystyle r+\left(\frac{1}{r}\right)^3$

and simplify, and if what I have done has no algebraic errors you should end up with $r=\pm i$

(note: in a competition you will get no marks for the final answer but just from the method you used to find it)

CB

3. I don't know if this is quite what is wanted but one can show that if $\left(r+ \frac{1}{r}\right)^2= 3$, then $r+ \frac{1}{r^3}= \frac{1}{r}$.

4. Hello, eric1299171!

Is there a typo?
If the second expression has $\,r^3$ twice,
. . there is a rather elegant solution.
And it looks more like a contest problem.

$\text{Given: }\:\left(r + \dfrac{1}{r}\right)^2 \,=\, 3.$
. . . . . . . .
$\text{Find: }\:r^3 + \dfrac{1}{r^3}$

We have: . . . . . $r + \dfrac{1}{r} \;=\;\sqrt{3}$

Then: . . . . . $\left(r + \dfrac{1}{r}\right)^3 \;=\;3\sqrt{3}$

. . . . $r^3 + 3r + \dfrac{3}{r} + \dfrac{1}{r^3} \;=\;3\sqrt{3}$

. . $\displaystyle r^3 + 3\underbrace{\left(r + \frac{1}{r}\right)}_{\text{This is }\sqrt{3}} + \frac{1}{r^3} \;=\;3\sqrt{3}$

. . . . . . $\displaystyle r^3 + 3\sqrt{3} + \frac{1}{r^3} \;=\;3\sqrt{3}$

Therefore: . . . . $r^3 + \dfrac{1}{r^3} \;=\;0$

5. I checked over the contest and the second expression is definetly r + 1/(r^3). But the answer key has zero. There might be a typo on a contest. Thanks for the reply! =)

6. Originally Posted by eric1299171
... Thanks for the reply! =)
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