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September 6th, 2008, 05:17 PM
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prove that  |
September 6th, 2008, 10:23 PM
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Hello, perash!
I got the first half . . .
I happened to notice that: .![\begin{array}{ccc}3 + 12\sqrt[3]{11} - 6\sqrt[3]{121} &=& \left(\sqrt[3]{11} - 2\right)^3 \\ \\[-3mm] 16 -27\sqrt[3]{11} + 9\sqrt[3]{121} &=& \left(3 - \sqrt[3]{11}\right)^3 \end{array}](http://www.mathhelpforum.com/math-help/latex2/img/79819210b6bd4d717439e2928080272c-1.gif "\begin{array}{ccc}3 + 12\sqrt[3]{11} - 6\sqrt[3]{121} &=& \left(\sqrt[3]{11} - 2\right)^3 \\ \\[-3mm] 16 -27\sqrt[3]{11} + 9\sqrt[3]{121} &=& \left(3 - \sqrt[3]{11}\right)^3 \end{array}")
(a) Therefore: .![A + B \;=\;\sqrt[3]{\left(\sqrt[3]{11}-2\right)^3} + \sqrt[3]{\left(3-\sqrt[3]{11}\right)^3} \;=\;\left(\sqrt[3]{11} - 2\right) + \left(3 - \sqrt[3]{11}\right) \;=\;1](http://www.mathhelpforum.com/math-help/latex2/img/99fbb20bee797f19f6d8aaf231f298bd-1.gif "A + B \;=\;\sqrt[3]{\left(\sqrt[3]{11}-2\right)^3} + \sqrt[3]{\left(3-\sqrt[3]{11}\right)^3} \;=\;\left(\sqrt[3]{11} - 2\right) + \left(3 - \sqrt[3]{11}\right) \;=\;1")
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September 7th, 2008, 07:06 PM
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In your final solution wouldnt the cube route and the 3rd pwr cancel out? |
September 7th, 2008, 07:14 PM
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Quote:
Originally Posted by sportsfan93b  In your final solution wouldnt the cube root and the 3rd pwr cancel out? | what do you mean? they did cancel out...
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September 7th, 2008, 07:14 PM
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What do you mean? It did cancel out in part A or did I misunderstand your question?
Lol I was too slow.
Last edited by 11rdc11; September 7th, 2008 at 07:16 PM.
Reason: I'm to slow
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![\text{Let: }\:A \;=\;\sqrt[3]{3 + 12\sqrt[3]{11} - 6\sqrt[3]{121}}\;\text{ and }\;B \;=\;\sqrt[3]{16 - 27\sqrt[3]{11} + 9\sqrt[3]{121}}](http://www.mathhelpforum.com/math-help/latex2/img/082d21188341f56c8e273a63f33b9657-1.gif "\text{Let: }\:A \;=\;\sqrt[3]{3 + 12\sqrt[3]{11} - 6\sqrt[3]{121}}\;\text{ and }\;B \;=\;\sqrt[3]{16 - 27\sqrt[3]{11} + 9\sqrt[3]{121}}")
