Question

A rationalizing factor of ( cube root of 9 - cube root of 3 + 1) options: a) cube root of 3 -1 b) cube root of 3 +1 c) cube root of 9 +1 d) cube root of 9-1

Answer 1

option b 3+1 answer of 9-cube root of 3+1

Answer 2

The rationalizing factor of $\sqrt[3]{9}-\sqrt[3]{3}+1$ is equal to (b) $\sqrt[3]{3} +1}$

Step-by-step Explanation

Given: An expression, $\sqrt[3]{9}-\sqrt[3]{3}+1$

To be found:

To find the rationalizing factor of the given expression.

Formula Used: Use the formula, $(x+1)(x^{2} -x+1)=x^{3}+1$    -------(1)

Solution:

We have the given expression, $\sqrt[3]{9}-\sqrt[3]{3}+1$

Rearranging the expression in exponential form, we get

$9^{\frac{1}{3} } -3^{\frac{1}{3} } +1\\\implies (3^{2}) ^{\frac{1}{3} } -3^{\frac{1}{3} } +1$         $(\because 9=3^{2} )$

Simplifying, we have

$3^{\frac{2}{3} } -3^{\frac{1}{3} } +1$  -------(2)

Now, comparing (2) and (3), take $x=3^{\frac{1}{3} }$ and substituting in (1), we get

$(3^{\frac{1}{3} }+1)((3^{\frac{1}{3} })^{2} -3^{\frac{1}{3} }+1)=(3^{\frac{1}{3} })^{3}+1\\\implies (3^{\frac{1}{3} }+1)(3^{\frac{2}{3} } -3^{\frac{1}{3} }+1)=3+1$

Equating to the form of (2), we get

$(3^{\frac{2}{3} } -3^{\frac{1}{3} }+1)=\frac{4}{3^{\frac{1}{3} }+1}\\\implies (3^{\frac{2}{3} } -3^{\frac{1}{3} }+1)=\frac{4}{\sqrt[3]{3} +1}$

A rationalizing factor rationalizes the given expression.

Here, $\sqrt[3]{3} +1}$ is the rationalizing factor for the given expression.

Therefore, from among the options, we get the correct answer as (b) $\sqrt[3]{3} +1}$

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