Question

Is there a simple way to find the zeros of x^6-1?

Answer 1

Let x³ = u

\(\Large u^2 = (x^3)^2 = x^6\)

u²-1 = 0

(u+1)(u-1)=0

u = ±1

x³ = ±1

x = ³√(±1) = ±1 = \(\boxed{\text{1 or -1}}\)

Answer 2

That's my approach. Is this clear?

Answer 3

That's a perfect solution @geerky42

@sfb Do you understand what geerky has done here?

Answer 4

No, I don't think so. I'm still trying to understand it.

Answer 5

What part don't you understand? I'd be glad to clarify it.

Answer 6

Okay, I got x=+/-1, x=0, x=+/-i

Answer 7

Would that be right?

Answer 8

x is equal only to ±1. Because if you plug in 0 and ±i into x⁶-1=0, it won't be equal to zero.

Answer 9

I originally got +/-1, but my teacher said I needed to find the 4 complex zeros.

Answer 10

@genius12 Can you help her? I'm not good with complex...

Answer 11

4 complex zeros for x^6 - 1? It only has 2 zeros both of which are complex. When I say they are complex I mean they are real numbers, but because real numbers are a subset of complex numbers, then any real roots are complex roots. What makes your teacher think that they're 4 solutions?

Answer 12

"These are the correct real zeros, but you need to find the 4 complex zeros.
Hint: factor (x6 - 1) into (x3 + 1)(x3 - 1), then keep factoring or solving to find the other complex zeros."

Answer 13

Oh lol. I see what you mean now.

Answer 14

I think I saw my mistakes, ³√(±1) is COMPLEX, not real. so far we have found two complex zeros.

Answer 15

According to Wolfram Alpha, x⁶-1 = (x-1) (x+1) (x²-x+1) (x²+x+1).

Answer 16

I would use the quadratic formula on (x^2-x+1) and (x²+x+1)?

Answer 17

That's how I got my i's.

Answer 18

"I would use the quadratic formula on (x^2-x+1) and (x²+x+1)?"

Yes.

Answer 19

Shoot. I screwed. Ya, do the factorisation like Wolframalpha shows and then use quadratic formula to factor the quadratics to get the complex roots.

Answer 20

Haha, okay. Thank you both for the help! It is much appreciated.

Answer 21

You're welcome.