Constructible Solutions We know that if a cubic equation with a rational coefficients has a constructible root, then the equation has a rational root.
Now let; $$x^3-2x+2\sqrt{2}=0$$
Could $\sqrt{2}t$ be a viable substitution?
 A: You might find this page interesting. It is proving that if $p(x)$ has no rational roots, then it does not have any constructible roots, but in the process it proves that if a cubic has a constructible root, then the equation has a rational root. I will write out their proof below.
We assume that $p(x)$ has a constructible root $r$ but no rational roots.
If we write $p(x)$ in factored form, then multiply the factors, then
$$p(x) = (x-r_1)(x-r_2)(x-r_3) = x^3-(r_1+r_2+r_3)x^2+(r_1 r_2+r_1 r_3+r_2 r_3)x-r_1 r_2 r_3$$
Thus, the coefficients $a, b, c$ are in terms of roots by:
$$a = (r_1+r_2+r_3)$$ $$b = (r_1 r_2+r_1 r_3+r_2 r_3)$$ $$c = -r_1 r_2 r_3$$
Because $p(x)$ is assumed to have a constructible root r but no rational roots, there is a shortest expanding sequence:  $Q, F[a(1)], , ……, F[a(n)]$ of square root extension fields such that $r$ is in $F[a(n)]$ but not in $Q$. Since $r$ is not in $Q$, $k$ must be a positive integer and:
$$r=p+q\sqrt{w}$$ Also, because $$(p-q\sqrt{w})(p+q\sqrt{w})=p^2-q^2w : p\ne0$$
Another root of $p(x) = 0$ is $$r'=p-q\sqrt{w}$$
The third root of the $p(x) = 0$ is given by $$r''=(-r)-r'=(-a)-2p$$
which is in the field $F[a(n)]$, contrary to the assumption that $F[a(1)], ... , F[a(n)]$ is the shortest sequence of expanding square-root extension fields that contains r. 
This contradicts the assumption that $p(x)$ has no rational roots but does have a constructible root. Therefore, the assumption that the cubic polynomial has a constructible root but no rational root is incorrect. We conclude that if the polynomial $p(x)$ has a constructible root, it must also have a rational root. Equivalently, if $p(x)$ has no rational roots, then its roots are not constructible numbers.
A: Let $x=\sqrt{2}\,t$. Then we are looking at the equation $t^3-3t+1=0$. 
