# [✓] Solve the following inequality with W|A?

Posted 3 months ago
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 Consider the following W|A input: (x^3+2x^2+3x)^(1/3)-x-1<=0 Solution Wolfram -> x>=0 ?!?!?!?!? why?Solution is : -inf <= x <= +inf x = -2 is ok!!! -2 <=0
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Posted 3 months ago
 Hello Faustino,I don't see that x = -2 is OK. For x = -2 the term from which you have to take root 1/3 is negative. The number you get after taking the root will be complex.
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Posted 3 months ago
 Just to illustrate: In[1]:= t = (x^3 + 2 x^2 + 3 x)^(1/3) - x - 1 Out[1]= -1 - x + (3 x + 2 x^2 + x^3)^(1/3) In[2]:= N[t /. x -> -2] Out[2]= 1.90856 + 1.57367 I 
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Posted 3 months ago
 No, with x=-2 -> -0,817<=0
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Posted 3 months ago
 Looks like you are assuming that 1 + (-6)^(1/3) is the same as 1 - 6^(1/3). It is not.
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Posted 3 months ago
 To indicate the issue noted by @RohitNamjoshi, here is the difference in results between using the exponent 1/3 vs using the real-valued CubeRoot function. Reduce[(x^3 + 2 x^2 + 3 x)^(1/3) - x - 1 <= 0, x] (* Out[366]= x >= 0 *) Reduce[CubeRoot[(x^3 + 2 x^2 + 3 x)] - x - 1 <= 0, x] (* Out[367]= x \[Element] Reals *) 
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Posted 3 months ago
 Sorry Daniel,your answer only showed up after I submitted my attempt at a reply.Cheers, Marco
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Posted 3 months ago
 You might want to look up Surd (https://reference.wolfram.com/language/ref/Surd.html). Especially, you might want to look at the "Possible Issues" section where it says:"On the negative real axis, Surd[x,n] is different from the principal root returned by Powerand look a the following examples. If you look at Surd instead of Power then 1 + Surd[-6,3] == 1 - Surd[6,3] is actually True. Rather than 1 + (-6)^(1/3) == 1 - 6^(1/3) which is False. Note that Reduce[Surd[x^3 + 2 x^2 + 3 x, 3] - x - 1 <= 0, x] gives the solution that you expect if I am not mistaken.Cheers,Marco
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