Question

which of the following explains why the given radical is not in simpliest form ^3√162

Answer 1

This is a cube root, so we look for factors of 162 which are perfect cubes.

Find the prime factors of 162:-

162 = 2 * 3 * 3 * 3* 3

27 = 3^3 is a perfect cube

162 = 6 * 27

so ^3√ 162 = ^3√6 * ^3√27 = ^3√6 * 3

so the simplest form is 3 ^3√6

Answer 2

It is not in simplest form since its simplest form is $3 \:. \:\:^3\sqrt{6}$

The given radical is $^3\sqrt{162}$.

Its simplification can be done if we factorize the number 162 to take out any factor which might be a perfect cube.

The simplification process can go like this below:

$\begin{aligned} ^3\sqrt{162} &= ^3\sqrt{2\times\ 3 \times 3 \times 3 \times 3}\\&= ^3\sqrt{2 \times 3^4}\\&= 3 \times ^3\sqrt{2\times 3}\\ &= 3 \times ^3\sqrt{6}\\\\&= 3 \:. \:\:^3\sqrt{6}\\\end{aligned}$

The simplified form thus is $3 \:. \:\:^3\sqrt{6}$

Since the given radical can be simplified even further, thus its not in simplest form.

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