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It works just the same as #14. Set x=<math>\sqrt[3]{2}+\sqrt[3]{4}</math> and cube both sides. With some factoring and recognizing that x=<math>\sqrt[3]{2}+\sqrt[3]{4}</math>, you get <math>x^3-6x-6</math> as the polynomial. | It works just the same as #14. Set x=<math>\sqrt[3]{2}+\sqrt[3]{4}</math> and cube both sides. With some factoring and recognizing that x=<math>\sqrt[3]{2}+\sqrt[3]{4}</math>, you get <math>x^3-6x-6</math> as the polynomial. | ||
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| + | I found this problem to be hard as well, but writing them as 2^(1/3), etc., helped a lot. | ||
Revision as of 02:20, 4 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?
--- See, I know that $ \sqrt[3]{4}=(\sqrt[3]{2})^2 $ but that didn't help me. I got #14 easily. What else is different about this problem?
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It works just the same as #14. Set x=$ \sqrt[3]{2}+\sqrt[3]{4} $ and cube both sides. With some factoring and recognizing that x=$ \sqrt[3]{2}+\sqrt[3]{4} $, you get $ x^3-6x-6 $ as the polynomial.
I found this problem to be hard as well, but writing them as 2^(1/3), etc., helped a lot.
