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A cube root is a number that, when multiplied by itself three times, gives the original number. For example, the cube root of 8 is 2 because

• 2 * 2 * 2 = 8

Cube roots are denoted by the symbol \( \sqrt[3]{\cdot} \). Unlike square roots, where we multiply twice, cube roots require three identical factors.
In rationalizing denominators, it's crucial to convert cube roots into whole numbers. This process makes calculations simpler and results cleaner.
With the fraction \(\frac{1}{\sqrt[3]{2}}\), the cube root of 2 in the denominator needs rationalization to eliminate the root. To do this, we multiply by the necessary form of cube roots that make the denominator a whole number. In this case, multiplying by \(\frac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}}\) does just that.

Fraction simplification aims to present fractions in their simplest forms, making them easier to work with. When simplifying, the goal is to reduce both the numerator and the denominator to the smallest possible terms.
The basic idea is to find a common factor and divide the numerator and the denominator by it. However, when dealing with cube roots, simplification involves rationalizing as well.

• This is why, after the step-by-step multiplication of \(\frac{1}{\sqrt[3]{2}}\) by \(\frac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}}\), we reached \(\frac{\sqrt[3]{2^2}}{2}\). Here, the denominator has been rationalized to a whole number 2.

Keep in mind that a fraction is considered simplified when both top and bottom no longer have common factors, and any roots are removed from the denominator.

Algebraic manipulation involves rearranging and transforming algebraic expressions and equations. This skill ensures an expression is more digestible or fits a desired form. This is key in rationalizing denominators, especially when roots are involved. For rationalizing \(\frac{1}{\sqrt[3]{2}}\), we apply algebraic manipulation by:

• Recognizing the need to alter the cubic root in the denominator to a whole number
• Using identity \(a^m \cdot a^n = a^{m+n}\) to simplify expressions.

Here, multiplying the denominator \(\sqrt[3]{2}\) by \(\sqrt[3]{2^2}\) results in a perfect cube \(2\), as \(\sqrt[3]{2}\cdot\sqrt[3]{4} = \sqrt[3]{8} = 2\).
Through manipulation, we transition the expression to an equivalent, more workable form. Understanding and practicing these techniques can greatly enhance problem-solving skills in algebra.