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When I type the following implicit equation into Mathematica I only get 'one quarter' of what the resulting plot should look like, i.e., I only get the upper-right quadrant.

ContourPlot[x^(2/3) + y^(2/3) == 1, {x, -1, 1}, {y, -1, 1}]

If you type $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1$ into this graphing calculator you will see the correct output which has a curve in each of the four quadrants.

So why is Mathematica giving an incorrect result, and is it possible to make it plot this implicit function correctly?

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marked as duplicate by rhermans, Bob Hanlon, Daniel Lichtblau, Henrik Schumacher, m_goldberg plotting Aug 8 '18 at 23:35

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    $\begingroup$ Try ContourPlot[Abs[x]^(2/3) + Abs[y]^(2/3) == 1, {x, -1, 1}, {y, -1, 1}]. Mathematica is giving the correct result; for instance, how to plot (-0.5)^(2/3) ? $\endgroup$ – b.gates.you.know.what Aug 8 '18 at 5:30
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    $\begingroup$ ContourPlot[CubeRoot[x^2]+CubeRoot[y^2]==1,{x,-1,1},{y,-1,1}] or ContourPlot[(x^2)^(1/3)+(y^2)^(1/3)==1,{x,-1,1},{y,-1,1}] $\endgroup$ – mathe Aug 8 '18 at 6:08
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    $\begingroup$ The calculator is apparently interpreting x^(a/b) as (x^a)^(1/b) whereas in Mathematica this is treated as (x^(1/b))^a. The two are of course not the same. The Mathematica treatment is a consequence of defining x^y as (in more customary notation) exp(y*log(x)) $\endgroup$ – Daniel Lichtblau Aug 8 '18 at 14:28
  • $\begingroup$ ContourPlot[(x^2)^(1/3) + (y^2)^(1/3) == 1, {x, -1, 1}, {y, -1, 1}] $\endgroup$ – Bob Hanlon Aug 8 '18 at 23:39

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