Thread: prove that
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Old September 6th, 2008, 10:23 PM
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Hello, perash!

I got the first half . . .

Quote:
\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}}


\text{Show (without using a calculator) that:}

. . (a)\;\;A + B \:=\:1\qquad\qquad (b)\;\;A\cdot B \:< \:\frac{1}{4}

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}


(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
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