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# Question

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.

Note by Sal Gard
2 months, 3 weeks ago

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Thanks. Sorry for inconvenience. · 2 months, 3 weeks ago

Why apologize? You did nothing wrong? We're here to learn.... · 2 months, 3 weeks ago

Thanks for the solution. I already had this but I was looking for the problem, not the answer. Thanks anyway. · 2 months, 3 weeks ago

ARML 1997 · 2 months, 3 weeks ago

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 $$\sqrt[3]{\dfrac49} + \sqrt[3]{-\dfrac29} + \sqrt[3]{\dfrac19}$$. · 2 months, 3 weeks ago