# How do I get a numerical cube root?

Posted 9 years ago
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 Hi Can't paste code on my machine.Input\.b2r = CubeRoot[((¹V)/Pi ) (3/4)];           (* need cube root of \.b2r 3Sqrt[\.b2r] *)Out put"\.b2r =" Surd[Quantity[5.46439*10^-69, ("Meters")^3], 3]I require a numerical value like 1.8 x 10^-23 metersPS 1V is a calculated volume, but any cube root will do.Many ThanksAndrew Pepes
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Posted 8 years ago
 Hi, I'm very new to Mathematica and I can't figure out how power simplifications are performed. If I try with N[Power[Quantity[4, ("Centimeters")^3]*3/4/Pi, 1/3], 3] I get 0.985 cm which is what I expect.But, if I try with something like Clear[r1, r2, v1, v2]; r1 = Quantity[2 , "Centimeters"] v1 = r1^3*3/4*Pi v2 = v1*5 r2 = N[CubeRoot[v2*3/4/Pi], 2] I get Surd[Quantity[45/2, "Centimeters"^3], 3] where I would expect CubeRoot["Centimeters"^3] to be simplified to "Centimeters".What am I doing wrong?Thank you for your help, Luca-
Posted 9 years ago
 Would something like this work?Power[Quantity[5.46439*10^-69, ("Meters")^3], 1/3]It givesQuantity[1.76136*10^-23, "Meters"](* 1.76136*10^(-23) m *)M.
Posted 9 years ago
 Hi MarcoBrilliant, I used it with my variables \.b2r = Power[Quantity[((¹V)/Pi ) (3/4)], 1/3];and it gives"\.b2r =" (Quantity[1.76136*10^-23, "Meters"])Many Thanks