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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!

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If a solution of $p(x) = x^3 -6x+2 \sqrt{\pi}$ was constructible, then

$$\pi= \frac{1}{2}\left(6x -x^3\right)^2$$

would be constructible as the square, the cube and the division by an integer of a constructible number is constructible.

But that can’t be as $\pi$ is transcendental while a constructible number is algebraic.

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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.

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