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Math  /  Algebra

Question5 Honors Question 1, 6.0.60\mathbf{6 . 0 . 6 0} Points: 0 of 1
Determine the smallest number both the numerator and denominator should be multiplied by to rationalize the denominator of the radical expression. 233\frac{2}{\sqrt[3]{3}}

Studdy Solution

STEP 1

1. We are given a radical expression with a cube root in the denominator.
2. To rationalize the denominator, we need to eliminate the cube root.
3. We will multiply both the numerator and the denominator by a number that will make the denominator a rational number.

STEP 2

1. Identify the form needed to rationalize the denominator.
2. Determine the smallest number to multiply both the numerator and the denominator by.
3. Rationalize the denominator by performing the multiplication.

STEP 3

Identify the form needed to rationalize the denominator. The denominator is 33\sqrt[3]{3}. To eliminate the cube root, we need to multiply by a form that results in a perfect cube.
The expression we need is (33)3(\sqrt[3]{3})^3, which is 33.

STEP 4

Determine the smallest number to multiply both the numerator and the denominator by. We need to multiply by (33)2(\sqrt[3]{3})^2 to make the denominator a perfect cube.
(33)2=93 (\sqrt[3]{3})^2 = \sqrt[3]{9}
Thus, the smallest number to multiply both the numerator and the denominator by is 93\sqrt[3]{9}.

STEP 5

Rationalize the denominator by performing the multiplication:
Multiply both the numerator and the denominator by 93\sqrt[3]{9}:
233×9393=2⋅93273 \frac{2}{\sqrt[3]{3}} \times \frac{\sqrt[3]{9}}{\sqrt[3]{9}} = \frac{2 \cdot \sqrt[3]{9}}{\sqrt[3]{27}}
Since 273=3\sqrt[3]{27} = 3, the expression becomes:
2⋅933 \frac{2 \cdot \sqrt[3]{9}}{3}
The denominator is now rationalized.

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