Question

The rationalizing factor of $ \sqrt[3]{{49}} $ is
A. $ \sqrt {49} $
B. $ \sqrt[3]{7} $
C. 0
D. None

Answer 1

Hint: Rationalizing a number means getting rid of any square roots or cube roots by multiplying the given number by its conjugate. The conjugate should be in such a form that when it is multiplied to the given number, the given number must become an integer. As we can see the given number $ \sqrt[3]{{49}} $ is a cube root of 49, which means it needs three similar numbers to get rid of that cube root. 49 can be written as a product of two 7s. So find the rationalizing factor such that $ \sqrt[3]{{49}} $ becomes an integer.

Complete step-by-step answer:
We are given to find the rationalizing factor of $ \sqrt[3]{{49}} $ .
49 can be written as seven times seven, which is 7 square.
 $ 49 = 7 \times 7 = {7^2} $
Therefore,
 $ \Rightarrow \sqrt[3]{{49}} $ becomes $ \sqrt[3]{{{7^2}}} $ .
To get rid of a cube root, we need a product of three similar numbers.
Here we have two 7s, and we need one more seven inside the cube root.
So we have to multiply $ \sqrt[3]{{49}} $ by $ \sqrt[3]{7} $ , because we will have one more seven inside the cube root.
This becomes,
 $ \Rightarrow \sqrt[3]{{{7^2}}} \times \sqrt[3]{7} = {7^{\dfrac{2}{3}}} \times {7^{\dfrac{1}{3}}} = {7^{\dfrac{2}{3} + \dfrac{1}{3}}} = {7^{\dfrac{3}{3}}} = {7^1} = 7 $
So when we multiply $ \sqrt[3]{{49}} $ by $ \sqrt[3]{7} $ , we will get an integer as a result.
Therefore, the rationalizing factor of $ \sqrt[3]{{49}} $ is $ \sqrt[3]{7} $
So, the correct answer is “Option B”.

Note: Whenever we send a variable or a number inside a square root, the number becomes its square such as 3 becomes 9, 5 becomes 25 etc; whenever we send a variable or a number inside a cube root, the number becomes its cube such as 3 becomes 27, 5 becomes 125 etc.