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I am trying to evaluate the cubic root of a negative number

$(5 (7 - 3 \sqrt{6}))^{1/3}$

by 

(5 (7 - 3 Sqrt[6]))^(1/3) // N

which returns only:

0.601656 + 1.0421 i

but 

NSolve[5 (7 - 3 Sqrt[6]) == x^3, x]

returns:

{{x -> -1.20331}, {x -> 0.601656 - 1.0421 i}, {x -> 0.601656 + 1.0421 i}}

Why does Mathematica not return the real root as default?

user6043040
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    That would be because, by convention, the principal cube root is defined as a complex number with positive imaginary part, which is what Mathematica returns. If you want the real solution, you can use Mathematica's CubeRoot function. – Pirx Nov 22 '16 at 03:35
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    Expanding on @Pirx 's comment, N will return the root that has the lowest angle when it is expressed in polar ($re^{i \theta}$) form. – JungHwan Min Nov 22 '16 at 04:00
  • Thank you! CubeRoot might be my choice for this case. Is there any built-in-function for 1/5, 1/7 power of negative numbers similar to CubeRoot for cubic root? – user6043040 Nov 22 '16 at 04:04
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    That would be Surd. – Brett Champion Nov 22 '16 at 04:11
  • Got it! Thank you for kind guidance! – user6043040 Nov 22 '16 at 04:22

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