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MathGroup Archive 2007

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Re: Factorizing...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74143] Re: Factorizing...
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Mon, 12 Mar 2007 22:07:35 -0500 (EST)
  • References: <esr0lj$6hp$1@smc.vnet.net>

On Mar 9, 9:00 am, "Bruno Campanini" <B... at gmail.com> wrote:
> How can I get it from Mathematica:
>
> Factor[2+SQR(5)] = ((1+SQR(5))/2)^3
>
> ???
>
> Bruno

In[5]:=
Off[N::meprec]
2 + Sqrt[5] == ((1 + Sqrt[5])/2)^3 //Simplify
On[N::meprec]
Out[6]=
True

In[10]:=
2 + Sqrt[5] == Expand[((1 + Sqrt[5])/2)^3]
Out[10]=
True

In[51]:=
RootReduce[((1 + Sqrt[5])/2)^3]
Out[51]=
2 + Sqrt[5]

In[54]:=
facrul = 2 + Sqrt[5] -> HoldForm[((1 + Sqrt[5])/2)^3]
Out[54]=
2 + Sqrt[5] -> HoldForm[((1/2)*(1 + Sqrt[5]))^3]

In[55]:=
2 + Sqrt[5] + Sqrt[2] + 3^(1/3) + Root[11 - 7*#1 + #1^5 & , 1] /.
facrul
Out[55]=
Sqrt[2] + 3^(1/3) + HoldForm[((1/2)*(1 + Sqrt[5]))^3] + Root[11 - 7*#1
+ #1^5 & , 1]

In[56]:=
ReleaseHold[%]
Out[56]=
Sqrt[2] + 3^(1/3) + (1/8)*(1 + Sqrt[5])^3 + Root[11 - 7*#1 + #1^5 & ,
1]


???



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