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October 15th, 2008, 05:26 PM
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| | cubic equation help show that one fo the roots of the equation  is  and find the other two roots | 
October 15th, 2008, 09:04 PM
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| | Quote:
Originally Posted by siegfried show that one fo the roots of the equation  is  and find the other two roots | Let  .
This gives,  .
Which simplifies to  .
Let
If  and  .
Then this means ![u=\sqrt[3]{A} = e^{2\pi i/9} u=\sqrt[3]{A} = e^{2\pi i/9}](http://www.mathhelpforum.com/math-help/latex2/img/f693cf496492bd6fcdb0f02b7adc89fd-1.gif) and ![v=\sqrt[3]{B} = e^{-2\pi i/9} v=\sqrt[3]{B} = e^{-2\pi i/9}](http://www.mathhelpforum.com/math-help/latex2/img/c69f8d4e0522f2d6d769ace89c8af08e-1.gif) would produce solutions:  (where
Simplification of these formulas leads to the solutions:  .
And so the three solutions to the original are: | | The following users thank ThePerfectHacker for this useful post: | |  | 
October 15th, 2008, 10:24 PM
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| | thank you theperfecthacker
i believe you are using cardanos method but i think there is an easier method than cardanos | 
October 15th, 2008, 10:30 PM
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Originally Posted by siegfried thank you theperfecthacker
i believe you are using cardanos method but i think there is an easier method than cardanos | It is Cardano's method with a lot of work omitted as well as computations. | 
October 16th, 2008, 05:07 AM
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| | Use the formulas  and  . These show that if  then  , and so  . But if  then  . That gives the equation  . The other two solutions will come from the other angles satisfying  , namely  and  . ( ThePerfectHacker had  , but I think  must be correct.) | | The Following 2 Users Say Thank You to Opalg For This Useful Post: | |  | 
October 16th, 2008, 09:25 AM
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Originally Posted by Opalg ( ThePerfectHacker had  , but I think  must be correct.) | Yes I made a mistake. But we are going to keep it secret.
The mistake was in the simplification. The solutions that I got to the depressed cubic are correct I just made mistakes in the identities.
Here is the corrected version:
1)
2)
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