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.

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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 $\sqrt[3]{\dfrac49} + \sqrt[3]{-\dfrac29} + \sqrt[3]{\dfrac19}$ .

Pi Han Goh - 4 years, 11 months ago

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

Sal Gard - 4 years, 11 months ago

ARML 1997

Pi Han Goh - 4 years, 11 months ago

Thanks. Sorry for inconvenience.

Sal Gard - 4 years, 11 months ago

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

Pi Han Goh - 4 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$