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Looking into the intersection of abstract algebra and geometry, it's well known that it is impossible to double the cube with ruler and compass, since $\sqrt[3]{2}$ is not constructible. However, I have read that if one is given a parabola in $\mathbb{R}^2$, it is indeed possible to double the cube. Out of curiosity, how is this done?

One thing I have noticed that given a parabola $y=(1/2)x^2$, then the circle centered on $(a,1)$ will meet the parabola at $2\sqrt[3]{a}$ as the x-coordinate of intersection, implying we can find $\sqrt[3]{a}$. Is it then possible to find $\sqrt[3]{2}$ in $\mathbb{R}^2$ if we're given any arbitrary parabola?

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Menaechmus's solution was to get the intersection of two parabolas sharing vertices, one with focal length (the length from focus to vertex) twice the other. The abscissa (or ordinate, depending on how you oriented your two parabolas in the Cartesian plane), should give a segment whose length is $\sqrt[3]{2}$ times the focal length of the parabola with shorter focal length. See for instance this book. –  J. M. Apr 14 '11 at 4:20

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The calculation that you made is very close to the proof you seek. Given a parabola, it is not hard to construct first of all its axis of symmetry, and then the focus and directrix. In particular we can construct axes so that the parabola has equation of the shape $y=ax^2$. We can construct $a$ since the focus is at distance $1/(4a)$ from the vertex.

Now construct the circle with center $(a^2,1/(2a))$ and passing through the origin. This meets the parabola at point with $x$-coordinate a root of the equation $$(x-a^2)^2 + (ax^2-1/(2a))^2=a^4 +1/(4a^2)$$ The equation simplifies to $a^2x^4 -2a^2x=0$. The roots are $0$ and $\sqrt[3]{2}$.

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There is also the solution that relies on finding the intersection of a parabola and hyperbola of appropriate dimensions... –  J. M. Apr 14 '11 at 4:25

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