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This question already has an answer here:

I want to solve a simple equation

$Assumptions = 0 < v < 1 && x ∈ Reals && v ∈ Reals;
eqn1 = v == 1 - (1 - x)^(3/2)
sol = Solve[eqn1, {x},Reals]

enter image description here

But if I do a few more steps by hand

eqn2 = (1 - v)^(2/3) == ((1 - x)^(3/2))^(2/3) // PowerExpand
mySol = Solve[eqn2, {x}]

enter image description here

Did I miss some assumptions or something else? The assumtions from above don't seem to affect Solve, only the domain parameter.

I know that this is not a proof!

Plot[{x /. mySol, x /. sol}, {v, 0, 1}, PlotStyle -> {Line, Dashed}]

enter image description here

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marked as duplicate by MarcoB, Feyre, Sascha, Mr.Wizard Jan 23 '17 at 12:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What are your assumptions on v? Mathematica assumes every variable is complex unless told otherwise. $\endgroup$ – J. M. is away Jan 20 '17 at 15:42
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    $\begingroup$ What about this Solve[eqn1, x, Reals] // ToRadicals // First ? $\endgroup$ – zhk Jan 20 '17 at 16:10
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It seems that you just want to avoid the complex roots, so,

eqn1 = v == 1 - (1 - x)^(3/2)
sol = Solve[eqn1, x, Reals] // ToRadicals // First // Simplify // 
PowerExpand // Simplify

enter image description here

Normal[sol]

enter image description here

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