constructible root from cubic polynomial

hanelliot

Does x^3 - 3x + 3sqrt(3) = have a constructible root?

my solution:
suppose a is a constructible root of the equation above.
we square both sides to get x^6 - 6x^4 + 9x^2 = 27.
since a is constructible, a^2 is constructible as well and we can turn this equation into cubic poly with rational coefficients, and it becomes y^3 - 6y^2 + 9y - 27 = 0.

If this cubic poly has a constructible root, it must have a rational root in form of m/n.
(m/n)^3 - 6(m/n)^2 + 9(m/n) = 27.

How do I proceed from here? Detailed steps would be appreciated..

robert Ihnot

The form will wind up as m^3-6m^2n+9mn^2-27 n^3 =0. This is supposed to be a solution. (We assume that m and n are relatively prime, that is, contain no common factors.) So look at m, there is only 1 term in which it is not present.

So take it from there!

HallsofIvy

If it has a rational root, m/n, then m must divide the constant term and n must divide the leading coefficient. Here the constant term is -27 so m must be 1, -1, 3, -3, 9, -9, 27, or -27. The leading coefficient is 1 so the n must be either 1 or -1 which means the only possible rational roots are 1, -1, 3, -3, 9, -9, 27, or -27. Do any of those satisfy the equation?

Count Iblis

You can apply the Rational Roots Theorem to

y^3 - 6y^2 + 9y - 27 = 0.

as HallsofIvy explained, but it is possible to reduce the number of root candidates by substitution.

Define

P(y) = y^3 - 6y^2 + 9y - 27

P(1) = -23, so y = 1 is not a root. Then, instead of trying out the other candidates, you can look at the polynomial:

Q(t) = P(1+t)

The coefficient of t^3 in Q(t) is 1 and the constant term in Q(t) is
Q(0) = P(1) = -23

This means that any rational root of Q(t) must be a divisor of 23, so the possible roots are:

t = 1, -1, 23, -23

The possible rational roots of P(y) are then the values of 1+t which are:

y = 2, 0, 24, -21

But any rational root must also be in the list give by HallsofIvy. Since none of the values for y listed above are on that list, there are no rational roots.