Which one of the following is the rationalizing factor of the number ? Question 57Answer a. b. c. d.

The rationalizing factor of the number $\frac{1}{\sqrt[3]{3}}$ is $\sqrt[3]{3}$.

Explanation:
A rationalizing factor of an irrational number is a factor that, when multiplied by the irrational number, results in a rational number.

Let the given number be $N = \frac{1}{\sqrt[3]{3}}$. We need to find an option $F$ such that $N \times F$ is a rational number.

Let's test each option:

• a. $\sqrt[3]{3}$ Multiply $N$ by $\sqrt[3]{3}$: $N \times \sqrt[3]{3} = \frac{1}{\sqrt[3]{3}} \times \sqrt[3]{3} = 1$ Since 1 is a rational number, $\sqrt[3]{3}$ is a rationalizing factor.

• b. $\sqrt[3]{9}$ Multiply $N$ by $\sqrt[3]{9}$: $N \times \sqrt[3]{9} = \frac{1}{\sqrt[3]{3}} \times \sqrt[3]{9} = \sqrt[3]{\frac{9}{3}} = \sqrt[3]{3}$ Since $\sqrt[3]{3}$ is an irrational number, $\sqrt[3]{9}$ is not a rationalizing factor for $N$.

• c. $\sqrt{3}$ Multiply $N$ by $\sqrt{3}$: $N \times \sqrt{3} = \frac{1}{\sqrt[3]{3}} \times \sqrt{3} = 3^{-1/3} \times 3^{1/2} = 3^{(1/2) - (1/3)} = 3^{(3-2)/6} = 3^{1/6} = \sqrt[6]{3}$ Since $\sqrt[6]{3}$ is an irrational number, $\sqrt{3}$ is not a rationalizing factor for $N$.

• **d. 3} Multiply $N$ by 3: $N \times 3 = \frac{1}{\sqrt[3]{3}} \times 3 = 3 \times 3^{-1/3} = 3^{1 - (1/3)} = 3^{2/3} = \sqrt[3]{9}$ Since $\sqrt[3]{9}$ is an irrational number, 3 is not a rationalizing factor for $N$.

Only option a, $\sqrt[3]{3}$, results in a rational number when multiplied by $\frac{1}{\sqrt[3]{3}}$.