Editing Doubling the cube (section)

==Solutions via means other than compass and straightedge==
Menaechmus' original solution involves the intersection of two [[conic]] curves. Other more complicated methods of doubling the cube involve [[neusis]], the [[cissoid of Diocles]], the [[Conchoid (mathematics)|conchoid of Nicomedes]], or the [[Philo line]]. [[Pandrosion]], a probably female mathematician of ancient Greece, found a numerically accurate approximate solution using planes in three dimensions, but was heavily criticized by [[Pappus of Alexandria]] for not providing a proper [[mathematical proof]].<ref>{{cite book|last=Knorr|first=Wilbur Richard|author-link=Wilbur Knorr|contribution=Pappus' texts on cube duplication|doi=10.1007/978-1-4612-3690-0_5|pages=[https://archive.org/details/textualstudiesin0000knor/page/63 63–76]|publisher=Birkhäuser|location=Boston|title=Textual Studies in Ancient and Medieval Geometry|url=https://archive.org/details/textualstudiesin0000knor|url-access=registration|year=1989|isbn=9780817633875 }}</ref> [[Archytas]] solved the problem in the 4th century BC using geometric construction in three dimensions, determining a certain point as the intersection of three surfaces of revolution.

Descartes' theory of geometric solution of equations uses a parabola to introduce cubic equations, in this way it is possible to set up an equation whose solution is a cube root of two. Note that the parabola itself is not constructible except by three dimensional methods.

False claims of doubling the cube with compass and straightedge abound in mathematical [[Crank (person)|crank]] literature ([[pseudomathematics]]).

Origami may also be used to construct the [[Mathematics of paper folding#Doubling the cube|cube root of two by folding paper]].

===Using a marked ruler===
[[File:Doubling the cube.svg|right|300px]]
There is a simple [[neusis construction]] using a marked ruler for a length which is the cube root of 2 times another length.<ref>{{cite book |title=100 Great Problems of Elementary Mathematics |first=Heinrich |last=Dörrie |publisher=Dover |year=1965 |page=171 |isbn=0486-61348-8|url=https://archive.org/details/100GreatProblemsOfElementaryMathematicsDoverHeinrichDrrie/page/171/mode/2up}}</ref>

#Mark a ruler with the given length; this will eventually be GH.
#Construct an equilateral triangle ABC with the given length as side.
#Extend AB an equal amount again to D.
#Extend the line BC forming the line CE.
#Extend the line DC forming the line CF.
#Place the marked ruler so it goes through A and one end, G, of the marked length falls on ray CF and the other end of the marked length, H, falls on ray CE. Thus GH is the given length.

Then AG is the given length times <math>\sqrt[3]{2}</math>.