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| − | As stated above: <math>\sqrt[3]{4}=(\sqrt[3]{2})^2</math> so | + | As stated above: <math>\sqrt[3]{4}=(\sqrt[3]{2})^2</math>, so [Q(<math>\sqrt[3]{2}+sqrt[3]{4}</math>:Q]=3. This means we need a polynomial of degree 3. You can use the same process as 14 by letting x=<math>\sqrt[3]{2}+sqrt[3]{4}</math> then find <math>\x^3</math> which I found to equal 6x-6. (I found it easier to refer to <math>\sqrt[3]{2}+sqrt[3]{4}</math> as <math>\2^(1/3)</math> and <math>\2^(2/3)</math> when doing the calculations.) |
I think this is right as the polynomial has <math>\sqrt[3]{2}+sqrt[3]{4}</math> as a zero. Did anyone else get this? | I think this is right as the polynomial has <math>\sqrt[3]{2}+sqrt[3]{4}</math> as a zero. Did anyone else get this? | ||
Revision as of 15:39, 3 December 2008
Following the same procedures as with a square root does not get me anywhere. Does anyone have any suggetions? on how to find the minimal polynomial for a cube root?! --Robertsr 23:49, 2 December 2008 (UTC)
Yea, I keep trying to solve this in a similar way as 14 but it doesn't work. Grrr.
Remember that $ \sqrt[3]{4}=(\sqrt[3]{2})^2 $ so $ \sqrt[3]{4} $ is the only important one
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As stated above: $ \sqrt[3]{4}=(\sqrt[3]{2})^2 $, so [Q($ \sqrt[3]{2}+sqrt[3]{4} $:Q]=3. This means we need a polynomial of degree 3. You can use the same process as 14 by letting x=$ \sqrt[3]{2}+sqrt[3]{4} $ then find $ \x^3 $ which I found to equal 6x-6. (I found it easier to refer to $ \sqrt[3]{2}+sqrt[3]{4} $ as $ \2^(1/3) $ and $ \2^(2/3) $ when doing the calculations.) I think this is right as the polynomial has $ \sqrt[3]{2}+sqrt[3]{4} $ as a zero. Did anyone else get this?
