Question

please step by step


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Answer 1

Required Answer:-

Here you can see a surd ($\sqrt[3]{9} - \sqrt[3]{3} + 1$) is given where the two of the terms are cube root of 9 and 3. To rationalize, we need to remove the cube root. This can be done by cubing it.

We know an algebraic identity:

$\boxed{ \rm{ {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2}) }}$

We can use this formula. Don't be confused, have a look on the next steps. Then u can understand, Why I used this identity to rationalize it.

Simplifying the surd:

$= \sqrt[3]{9} - \sqrt[3]{3} + 1$

$= \sqrt[3]{ {3}^{2} } - \sqrt[3]{3} + {1}^{2}$

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

Let a = $\sqrt[3]{3}$ and b = 1. Then the above is now in the form a² - ab + b². To rationalize, we can use the above identity. The other pair will be the rationalizing factor i.e. a + b. Hence, the rationalizing factor is:

$= \sqrt[3]{3} + 1$

Hence,

• The correct option is B) = $\sqrt[3]{3} + 1$ which is rationalizing factor of above.

• If we multiply these two, $(\sqrt[3]{3} + 1)(\sqrt[3]{9} - \sqrt[3]{3} + 1)$, it gives a³ + b³ i.e. 3 + 1 = 4

Answer 2

Step-by-step explanation:

Given:-

$\tt{The \: rationalising \: factor \: of \: \sqrt[3]{9} - \sqrt[3]{3} + 1) \: is}$

$(A) \: \sqrt[3]{3 } - 1$

$(B) \: \sqrt[3]{3 } + 1$

$(C) \: \sqrt[3]{9 } + 1$

$(D) \: \sqrt[3]{9} - 1$

Correct answer:-

$(B) \: \sqrt[3]{3} + 1 \: ✔$

Description of correct answer:-

$\sqrt[3]{9} - \sqrt[3]{3} + 1 = (3 {)}^{ \frac{2}{3} } - (3 {)}^{ \frac{1}{3} } + (1 {)}^{ \frac{2}{3} }$

$∴ \: ( \sqrt[3]{3} + 1)( \sqrt[3]{9} - \sqrt[3]{3} + 1) = ( {3}^{ \frac{1}{3} } {)}^{3} + 1$

$= 3 + 1 = 4$

$[∴ {a}^{3} + {b}^{3} = (a + b) = ( {a}^{2} - ab + {b}^{2} )]$

$\large \: ∴ \: Rationalising \: factor= \sqrt[3]{3} + 1$

I hope it's help you..☺