What is the cube root of 297?

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What is the cube root of 297?

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Answer 1 · 2015-09-21T23:14:26

$\sqrt[3]{297} = 3 \sqrt[3]{11}$ $root(3)(297) = 3root(3)(11)$

Explanation:

$\sqrt[3]{297}$ $root(3)(297)$

Factor it out, sice $2 + 9 + 7 = 18$ $2+9+7 = 18$ we know 297 is divisible by 9.

$\sqrt[3]{297} = \sqrt[3]{33 \cdot 9}$ $root(3)(297) = root(3)(33*9)$

Since $3 + 3 = 6$ $3+3 = 6$ we know 33 is divisible by 3

$\sqrt[3]{33 \cdot 9} = \sqrt[3]{11 \cdot 3 \cdot 9}$ $root(3)(33*9) = root(3)(11*3*9)$

11 is a prime number so there's no more factoring do. We know that $9 = 3^{2}$ $9 = 3^2$ , so we can rewrite $3 \cdot 9 = 3 \cdot 3^{2} = 3^{3}$ $3*9 = 3*3^2 = 3^3$

$\sqrt[3]{33 \cdot 9} = \sqrt[3]{11 \cdot 3^{3}}$ $root(3)(33*9) = root(3)(11*3^3)$

The $3^{3}$ $3^3$ can go out of the root, so

$\sqrt[3]{297} = 3 \sqrt[3]{11}$ $root(3)(297) = 3root(3)(11)$

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What is the cube root of 297?

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