I thought up a new approach to the problem about \(\large \sqrt[3]{\sqrt[3]{2} - 1} \) and its simplifications. I know Calvin posted this a while back and I was wondering if any of you had a link so I can answer it.

Rationalize numerator by multiplying the expression by \( \dfrac{\sqrt[3]{\sqrt[3]{4} + \sqrt[3]{2} + 1}}{\sqrt[3]{\sqrt[3]{4} + \sqrt[3]{2} + 1}} \) then simplify the expression, then multiply it by \( \dfrac{\sqrt[3]{3}}{\sqrt[3]{3}} \), you get \( \dfrac{ \sqrt[3]{3}}{1 + \sqrt[3]{3}} \). Finally, rationalize the denominator to obtain \( \sqrt[3]{\dfrac49} + \sqrt[3]{-\dfrac29} + \sqrt[3]{\dfrac19} \).

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TopNewestThanks. Sorry for inconvenience.

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Why apologize? You did nothing wrong? We're here to learn....

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Thanks for the solution. I already had this but I was looking for the problem, not the answer. Thanks anyway.

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ARML 1997

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Rationalize numerator by multiplying the expression by \( \dfrac{\sqrt[3]{\sqrt[3]{4} + \sqrt[3]{2} + 1}}{\sqrt[3]{\sqrt[3]{4} + \sqrt[3]{2} + 1}} \) then simplify the expression, then multiply it by \( \dfrac{\sqrt[3]{3}}{\sqrt[3]{3}} \), you get \( \dfrac{ \sqrt[3]{3}}{1 + \sqrt[3]{3}} \). Finally, rationalize the denominator to obtain \( \sqrt[3]{\dfrac49} + \sqrt[3]{-\dfrac29} + \sqrt[3]{\dfrac19} \).

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