# Constructible roots

1. Apr 8, 2009

1. The problem statement, all variables and given/known data

prove that x^(6) - x^(2) +2 =0 has no constructible roots

2. Relevant equations

see above

3. The attempt at a solution

I have to divide the equation by x^(3) which would give me x^(3) - x^(-1) + 2x^(-2)= 0
I cant find a suitable substitution in terms of x which would change this into a proper cubic equation and see if it has rational roots.

2. Apr 8, 2009

### Billy Bob

Typo? It has no real roots at all.

3. Apr 9, 2009

how do i show it then? by assuming x=p/q as a rational root and then going on from there?

4. Apr 9, 2009

ok i made the substitution x=y+1 and now I have this polynomial y^6 + 6y^5 + 15y^4 + 20y^3 + 14y^2 + 4y + 2 = 0, all i have to do is show that this has irrational roots but i just dont know how.

5. Apr 9, 2009

### Hurkyl

Staff Emeritus
What precisely do you mean by "constructible"? In the version I'm familiar with, many complex numbers are constructible. (In particular, anything with constructible real and imaginary parts is constructible)

6. Apr 9, 2009

anything that can be drawn using a straight edge and a compass, the only problem i have is to show that the equation i have written has no rational roots.

7. Apr 9, 2009

### Billy Bob

No, you have to use a cubic for the theorem to apply.

Why would irrational roots tell you anything about constructible roots? Many irrational numbers, for example $\sqrt{2}$, are constructible.
It's hard to tell you how to proceed without know what fact, theorems, etc. you have to work with. If it were me, I would just use the fact that all constructible numbers are algebraic of order a power of 2. Then show that none of the roots of $x^6- x^2+ 2= 0$ (which is equivalent to $y^3- y+ 2= 0$ with $y= x^2$) are algebraic of order 2 or 0.