Arrange √(2), [3]3 and [6]6 in increasing order.

Question 34

HardOriginal question

Arrange $\sqrt{2}$ , $\sqrt[3]{3}$ and $\sqrt[6]{6}$ in increasing order.

\sqrt{2} < \sqrt[3]{3} < \sqrt[6]{6}

Correct answerYour answer

\sqrt[6]{6} < \sqrt[3]{3} < \sqrt{2}

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\sqrt[6]{6} < \sqrt{2} < \sqrt[3]{3}

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\sqrt{2} < \sqrt[6]{6} < \sqrt[3]{3}

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CorrectNot quiteAnswer C

Correct answer: C

Raise each to the sixth power, which keeps the order of positive numbers: $(\sqrt{2})^{6} = 8$ , $(\sqrt[3]{3})^{6} = 9$ , $(\sqrt[6]{6})^{6} = 6$ .
$6 < 8 < 9$ , so $\sqrt[6]{6} < \sqrt{2} < \sqrt[3]{3}$ .

Worth noting

$\sqrt[6]{6}$ has the largest number under the root, but the sixth root shrinks it the most.

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