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  "body": "\n## Proof\n\nLet $c = \\sqrt[3]{2}$, the real cube root of $2$, and let $\\mathcal{C}$ be the set of the [[constructible numbers|0FMN2MN7TM]].\n\n1. The polynomial $f = x^3 - 2$ satisfies [[Eisenstein's criterion|02BPKD7TMD]] with $p = 2$: $2$ divides $0, 0, -2$, $2$ does not divide $1$, and $4$ does not divide $-2$. So $f$ is irreducible in $\\Q[x]$.\n2. The [[minimal polynomial|0GK1WB9R98]] of $c$ over $\\Q$ [[divides|06K7SARZKR]] $f$, since $f(c) = 0$. As $f$ is irreducible and both are monic, the minimal polynomial is $f$.\n3. By the [[degree of a simple extension|0N5QW9K7MH]], $[\\Q(c) : \\Q] = 3$.\n4. If $c$ were constructible, then $[\\Q(c) : \\Q]$ would be a [[power of $2$|0SSSM1D475]]. But $3$ is not a power of $2$. So $c \\notin \\mathcal{C}$.\n\nNow we turn to the construction problem. Take the edge of the given cube as the segment from $0$ to $1$. A [[doubling of the cube|04SD3F4FCR]] would construct two points $P, Q$ with $\\abs{P - Q} = c$, since the doubled cube has volume $2 = c^3$.\n\n5. By the [[field theorem|0PN9WMRWT7]], $\\mathcal{C}$ is a subfield of $\\C$ closed under square roots, and it is the smallest such subfield.\n6. Let $\\overline{\\mathcal{C}} = \\set{\\bar z : z \\in \\mathcal{C}}$, using the [[complex conjugate|0BJV24C4SY]]. Conjugation preserves sums, products and inverses, so $\\overline{\\mathcal{C}}$ is a subfield of $\\C$.\n7. If $w = \\bar z$ with $z \\in \\mathcal{C}$, and $s^2 = w$, then $\\bar s^2 = z$. So $\\bar s \\in \\mathcal{C}$ and $s \\in \\overline{\\mathcal{C}}$. Thus $\\overline{\\mathcal{C}}$ is closed under square roots.\n8. By minimality, $\\mathcal{C} \\subseteq \\overline{\\mathcal{C}}$. So for $z \\in \\mathcal{C}$ we have $z = \\bar u$ with $u \\in \\mathcal{C}$, and $\\bar z = u \\in \\mathcal{C}$.\n9. Let $w = P - Q$. Then $w \\in \\mathcal{C}$, so $\\bar w \\in \\mathcal{C}$ by step 8, and [[$w \\bar w = \\abs{w}^2$|0ZB7J071WQ]] lies in $\\mathcal{C}$.\n10. The [[modulus|0KW2X0TFBX]] $\\abs{w} = c$ is a square root of $\\abs{w}^2$, so $c \\in \\mathcal{C}$ by step 5. This contradicts step 4.\n11. So $\\sqrt[3]{2}$ is not constructible, and the cube cannot be doubled with straightedge and compass.\n\n## Idea\n\nThe number $\\sqrt[3]{2}$ has degree $3$ over $\\Q$, while constructible numbers have degree a power of $2$. A construction of the doubled edge would make its length constructible, because the constructible numbers are closed under conjugation, products and square roots.\n",
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