       Re: Cube root of -1 and 1

• To: mathgroup at smc.vnet.net
• Subject: [mg91028] Re: Cube root of -1 and 1
• From: "David Park" <djmpark at comcast.net>
• Date: Mon, 4 Aug 2008 03:22:23 -0400 (EDT)
• References: <g6h5fm\$h26\$1@smc.vnet.net>

```Mathematica just does a straight forward calculation and does not try to
find multiple roots.

(Sqrt - 7)^(1/3) // Trace
{{{Sqrt,6},6-7,-1},{{1/3,1/3},1/3,1/3},(-1)^(1/3)}

(Sqrt-5)^(1/3)//Trace
{{{Sqrt,6},6-5,1},{{1/3,1/3},1/3,1/3},1^(1/3),1}

(Notice that 1^x gives 1.)

If you want all the roots use Solve:

Solve[w^3 == (Sqrt - 7)]
{{w->-1},{w->(-1)^(1/3)},{w->-(-1)^(2/3)}}

Solve[w^3 == (Sqrt - 5)]
{{w->1},{w->-(-1)^(1/3)},{w->(-1)^(2/3)}}

--
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/

"Bob F" <deepyogurt at gmail.com> wrote in message
news:g6h5fm\$h26\$1 at smc.vnet.net...
> Could someone explain why Mathematica evaluates these so differently?
>
> In:=
>
> (Sqrt - 7)^(1/3)
> (Sqrt - 5)^(1/3)
>
> Out= (-1)^(1/3)
>
> Out= 1
>
> In other words why isn't (-1)^1/3 expressed as -1 ??
>
> Thanks...
>
> -Bob
>

```

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