For anyone who's adventurous enough....

If a is a complex number and

$$x ^{10}+x ^{5}+1=0$$

Find

$$x ^{2005}+1/x ^{2005}$$

Question

For anyone who's adventurous enough....

If a is a complex number and

$$x ^{10}+x ^{5}+1=0$$

Find

$$x ^{2005}+1/x ^{2005}$$

anonymous · Answer

Also, second line should read "If x is a complex number"...sorry about the typo.

jamesj · Answer

hence $$ x^5 = \frac{-1 \pm \sqrt{3} i}{2} = e^{\pm \pi i/6} $$

jamesj · Answer

Now it becomes very easy to calculate x^2005 and x^-2005

jamesj · Answer

oops, wrong argument for exp:
$$ x = e^{2\pi i /3}  \ \ \ or \ \ \ x = e^{4 \pi i /3} $$

anonymous · Answer

So what are you getting as a final answer?....My approach is rather different from yours, but maybe both work?

jamesj · Answer

I'd rather not calculate it, but I know it's a real number.

anonymous · Answer

A real number, yes...and a fairly simple one at that...I'll post my solution in a few minutes...

anonymous · Answer

Here's my approach, which I think is much simpler...

If we multiply the original equation by $$x ^{5}$$

We get

$$x ^{15}+x ^{10}+x ^{5}=0$$ 

If we subtract the original equation from the above equation we get


$$x ^{15}=1$$

Since x cannot be 0 (as it would yield 1=0 in the original equation), dividing 

$$x ^{10}+x ^{5}+1=0  $$

by

$$x ^{5}$$

yields 

$$x ^{5}+1/x ^{5}=-1$$

and squaring that gets


$$x ^{10}+1/x ^{10}=-1$$

Now,


$$x ^{2005}+1/x ^{2005}=(x ^{15})^{133}x ^{10}+1/(x ^{15})^{133}x ^{10}$$

=$$x ^{10}+1/x ^{10}=-1$$

Your way will likely give the right answer as well. Medal for your attempt.

jamesj · Answer

Not "likely", will. :-)