The real cubic root expression

Question

$2x^3-2x^2-3x+2=0$ has 3 real root, but they are all express in such way:

$x=\dfrac{1}{3}\left(1+\dfrac{\sqrt[3]{-23+3i\sqrt{237}}}{\sqrt[3]{2^2}}+\dfrac{11}{\sqrt[3]{2(-23+3i\sqrt{237}})}\right)$

it is in complex format.

there is a way like this:

$x=\dfrac{1}{3}+\dfrac{\sqrt{22}}{3}\cos{\dfrac{\arccos{\dfrac{-23}{11\sqrt{22}}}}{3}}$

it is in real format.

but if I want to explain to some one that the root like this can't be constructed by straight and compass,I had problem.

if the root $x$ is shown in cubic root ,such as $\sqrt[3]{2+\sqrt{23}}$, then it is easy to say. but for a angle of $\dfrac{\alpha}{3}$, there is some possibility to construct depends on $\alpha$.

I try to convert $\dfrac{\sqrt{22}}{3}\cos{\dfrac{\arccos{\dfrac{-23}{11\sqrt{22}}}}{3}}$ into $\sqrt[3]{p+\sqrt{q}}$ but I fall into a loop of cubic root so I can't find such a way.

Is it means that there is no possibility to find such expression ?

then how can I explain this root can't be constructed by straight and compass?

Answer 1

The only lengths that can be constructed by ruler and compass are those which are algebraic numbers whose degree is a power of $2$.

Solutions of your cubic will have degree $3$, which is not a power of $2$, so they are not constructible.

How do we know they have degree $3$ and not less? - because the cubic is irreducible, that is, it cannot be factorised into polynomials of smaller degree, having integer or rational coefficients.

How do we know it is irreducible? - since it is cubic, if it has (non-trivial) factors at all, it must have a linear factor, and therefore it must have a rational root. But it doesn't.

How do we know it doesn't have a rational root? - if it had a root say $p/q$, where $p,q$ are integers with no common factor, then substituting $x=p/q$ and multiplying by $q^3$ gives $$2p^3-2p^2q-3pq^2+2q^3=0\ .$$ This shows that $p$ and $q$ must both be factors of $2$, so they must be $\pm1$ or $\pm2$. This gives $$x=\pm\frac{1}{2}\,,\ \pm\frac{1}{1}$$ and a couple more as the only possible rational roots; but checking, none of them works, so there are no rational roots.

Answer 2

Field theory has a straightforward answer to this:

If $\alpha$ is a constructible number, then $\mathbb{Q}[\alpha]$ is a field extension over the rationals with degree that is a power of $2$. Now if you can show that the polynomial is irreducible over $\mathbb{Q}$, perhaps using the Rational Roots theorem, then it will be the case that $[\mathbb{Q}[\alpha]:\mathbb{Q}] = 3$, which is not a power of $2$. Therefore, $\alpha$ is not constructible as a length.

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Side note/warning: The converse is NOT true. If $[\mathbb{Q}[\alpha]:\mathbb{Q}] = 2^k$, then it is possible that $\alpha$ isn't constructible.

Answer 3

I can be wrong but I have the feeling that the solutions you give may be not correct. For $$2x^3-2x^2-3x+2=0$$ the solutions expressed as complex numbers are $$x_1=\frac{1}{3} \left(1+\frac{\sqrt[3]{-23+3 i \sqrt{237}}}{2^{2/3}}+\frac{11}{\sqrt[3]{2 \left(-23+3 i \sqrt{237}\right)}}\right)$$ $$x_2=\frac{1}{3}-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{-23+3 i \sqrt{237}}}{6\ 2^{2/3}}-\frac{11 \left(1-i \sqrt{3}\right)}{6 \sqrt[3]{2 \left(-23+3 i \sqrt{237}\right)}}$$ $$x_3=\frac{1}{3}-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{-23+3 i \sqrt{237}}}{6\ 2^{2/3}}-\frac{11 \left(1+i \sqrt{3}\right)}{6 \sqrt[3]{2 \left(-23+3 i \sqrt{237}\right)}}$$ which simplify to $$x_1=\frac{1}{3}+\sqrt{\frac{11}{6}} \sin \left(\frac{1}{3} \tan ^{-1}\left(\frac{3 \sqrt{237}}{23}\right)\right)+\frac{1}{3} \sqrt{\frac{11}{2}} \cos \left(\frac{1}{3} \tan ^{-1}\left(\frac{3 \sqrt{237}}{23}\right)\right)$$ $$x_2=\frac{1}{3}-\sqrt{\frac{11}{6}} \sin \left(\frac{1}{3} \tan ^{-1}\left(\frac{3 \sqrt{237}}{23}\right)\right)+\frac{1}{3} \sqrt{\frac{11}{2}} \cos \left(\frac{1}{3} \tan ^{-1}\left(\frac{3 \sqrt{237}}{23}\right)\right)$$ $$x_3=\frac{1}{3}-\frac{1}{3} \sqrt{22} \cos \left(\frac{1}{3} \tan ^{-1}\left(\frac{3 \sqrt{237}}{23}\right)\right)$$