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To: King Prout

x**3-1=0 is the equation we need to solve. -1 is one solution.

(x**3-1)/(x-1)=x**2+x+1 so we solve this by the quadratic formula.

x=(-1+Sqrt(-3))/2 or (-1-Sqrt(-3))/2

These are the others cube roots. Multiplying by 3 gives the cube roots of (-27).


59 posted on 12/04/2005 10:12:43 AM PST by Doctor Stochastic (Vegetabilisch = chaotisch ist der Charakter der Modernen. - Friedrich Schlegel)
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To: Doctor Stochastic
(x**3-1)/(x-1)=x**2+x+1 so we solve this by the quadratic formula.

x=(-1+Sqrt(-3))/2 or (-1-Sqrt(-3))/2

These are the others cube roots. Multiplying by 3 gives the cube roots of (-27).

Does sqrt(-1) have two roots?

62 posted on 12/04/2005 10:19:17 AM PST by bobdsmith
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To: Doctor Stochastic

inquiry: does x**3 = x^3?


67 posted on 12/04/2005 10:31:33 AM PST by King Prout (many accuse me of being overly literal... this would not be a problem if many were not under-precise)
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To: Doctor Stochastic
x**3-1=0 is the equation we need to solve. -1 is one solution.

I thought we were discussing the cube root of negative one.
If this is so, wouldn't it be "x**3 + 1 = 0"?

72 posted on 12/04/2005 10:37:17 AM PST by King Prout (many accuse me of being overly literal... this would not be a problem if many were not under-precise)
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