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This question already has an answer here:

If in WolframAlpha I put (-8)^(2/3) then it outputs the main root and 3 roots solution too (to the end in a polar graphic). How can I get the 3 root solution of (-8)^(2/3) in Mathematica Desktop?
I tried with: Solve[(x^(3/2) + 8 == 0), VerifySolutions -> False] but the output is {{x -> -2 + 2 I Sqrt[3]}} ... I don't understand, help me please.

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marked as duplicate by J. M. is away Oct 21 '18 at 15:26

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  • $\begingroup$ NSolve[x^(3/2) + 8 == 0, x] gives two solutions {{x -> -2. + 3.4641 I}, {x -> -2. - 3.4641 I}} $\endgroup$ – Ulrich Neumann Sep 20 '18 at 15:17
  • $\begingroup$ thanks for you request, but the real root is not here, I'd like in algebraic form too. $\endgroup$ – Javier Giménez Moya Sep 20 '18 at 15:19
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    $\begingroup$ Maybe In[921]:= Solve[x^3 == (-8)^2, x] Out[921]= {{x -> 4}, {x -> -4 (-1)^(1/3)}, {x -> 4 (-1)^(2/3)}} ?? $\endgroup$ – Daniel Lichtblau Sep 20 '18 at 15:22
  • $\begingroup$ @ Javier Giménez Moya Ok, but the number of solutions might find your attention $\endgroup$ – Ulrich Neumann Sep 20 '18 at 15:23
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    $\begingroup$ The real "root" is not a valid solution of x^(3/2) + 8 == 0, so the WolframAlpha polar plot output is misleading at best. $\endgroup$ – Carl Woll Sep 20 '18 at 15:24
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One approach is to use Reduce:

Reduce[x^(3/2) + 8 == 0]
x == -4 (-1)^(1/3) || x == 4 (-1)^(2/3)

or you can get it in more familiar form using ExpToTrig

Reduce[x^(3/2) + 8 == 0] // ExpToTrig
x == -2 - 2 I Sqrt[3] || x == -2 + 2 I Sqrt[3]
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By plotting the real and imaginary part separately

z^(3/2) + 8 == 0 /. z -> (x + I y) // ReIm // ComplexExpand
ContourPlot[Evaluate@%, {x, -5, 5}, {y, -5, 5}]

Plot of solutions in complex plane

you can see that, as Carl Woll writes in the comments, only the solutions which don't lie on the real line have both, vanishing real and imaginary part. The solution candidate on the real line at x==4 only has a vanishing imaginary part but real part 16.

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  • $\begingroup$ I thougth that if Power[(-8)^2, (3)^-1] was 4, one root for (-8)^(2/3) was 4 too. $\endgroup$ – Javier Giménez Moya Sep 20 '18 at 15:38
  • $\begingroup$ @JavierGiménezMoya Squaring the -8 adds a new solution, because we cannot distinguish anymore if we squared 8 or -8. Basically you have to decide from your original problem which you are trying to solve what is the correct equation to solve here. Either z^(3/2) == -8 or z^3 == 64. The first one having the two solutions with imaginary part only and the second equation giving you all three roots of unity (z^3==1) scaled by four. $\endgroup$ – Thies Heidecke Sep 21 '18 at 11:30
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If you are interested in the non-principal roots, first note that:

With[{r = x^(2/3)}, r^(3/2)]

x

So, letting x=-8 we see that:

r^(3/2) == -8

Hence the non-principal roots can be determined with:

Reduce[r^(3/2) + 8 == 0, r]

r == -4 (-1)^(1/3) || r == 4 (-1)^(2/3)

Notice that r == 4 is not a root of the above equation. The WA polar root plot has acquired a parasitic (i.e. false) root.

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  • $\begingroup$ I thougth that if Power[(-8)^2, (3)^-1] was 4, then almost one root for (-8)^(2/3) should be 4 too. (alpha wolfram show it) $\endgroup$ – Javier Giménez Moya Sep 20 '18 at 15:42

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