# Prove that the following equation has no constructible solutions.

Prove that the following equation has no constructible solution: $$\ x^3 - 6x + 2\sqrt{\pi} = 0$$

The way I am trying to approach is that: I want to transform the equation into some integer coefficient equation and use the Rational Root Theorem to prove that the corresponding equation has no rational root, then by Theorem: if a cubic polynomial with rational coefficients has a constructible root, then it must also have a rational root. (using contrapositive) to conclude that the original equation has no constructible solution.

However, I get stuck since $$\sqrt{\pi}$$ is not a constructible number. Therefore I can't come up with a corresponding integer coefficients equation and use the Rational Root Theorem to proceed.

Can you please point me in the right direction. Thanks in advance!

If a solution of $$p(x) = x^3 -6x+2 \sqrt{\pi}$$ was constructible, then
$$\pi= \frac{1}{2}\left(6x -x^3\right)^2$$
But that can’t be as $$\pi$$ is transcendental while a constructible number is algebraic.
Suppose that that equation has a constructible root $$r$$.
Then $$r^3-6r+2\sqrt{\pi}=0$$, or $$\sqrt{\pi}=\frac{6r-r^3}{2},$$ which is in $$\Bbb{Q}(r)$$, contradicting that $$\sqrt{\pi}$$ isn't constructible.