Question

what is the rationalising factor of ⁴√343 pls explain

Answer 1

$\sqrt[4]{343} = ({343})^{ \frac{1}{4} } \\ = {(( {7)}^{3} })^{ \frac{1}{4} } \\ = {7}^{ \frac{3}{4} }$
rationalize factor is calculated when this term is in denominator.
for this RF is same 7^3/4

Answer 2

Concept

A number that is needed to multiply in order to make an irrational number as rational number is known as rationalizing factor. The expression that is multiplied with an irrational expression to obtain a rational number is called the "rationalizing factor".

Given

Number=$\sqrt[4]{343}$

To find

We have to find the rationalzing factor of  $\sqrt[4]{343}$.

Explanation

To find the rationalizing factor of $\sqrt[4]{343}$ we need to first break 343 in order  of cube which is 7 means 7*7*7.

$\sqrt[4]{343}$=$\sqrt[4]{7^{3} }$

so the rationalizing factor is calculated above.

Hence the rationalizing factor of $\sqrt[4]{343}$=$7^{1/4}$ because it converts the irrational number into rational number.

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