Proving No Constructible Roots for x^(6) - x^(2) + 2 = 0

[saadsarfraz]

Homework Statement

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

Homework Equations

see above

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 can't find a suitable substitution in terms of x which would change this into a proper cubic equation and see if it has rational roots.

 

[Billy Bob]

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

Typo? It has no real roots at all.

 

[saadsarfraz]

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

 

[saadsarfraz]

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 don't know how.

 

[Hurkyl]

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)

 

[saadsarfraz]

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.

 

[Billy Bob]

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 don't know how.

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

How about w=x^2?

 

[HallsofIvy]

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.