Properties of/Simplifying Radicals

[lolily]

Hi! So, on my homework, I came across the problem ∛54/∛10. I didn't understand the method that the teacher had given to solve this problem and I plugged it into a Photomath calculator, and the first step simplified the expression to ∛27/∛5. The eventual final answer was 3∛25/5, which is the correct answer listed in the back of the book, but I'm extremely confused as to how to get ∛27/∛5 from ∛54/∛10. Can anyone explain that first step to me? Thanks a bunch in advance.

 

[tkhunny]

Hi! So, on my homework, I came across the problem ∛54/∛10. I didn't understand the method that the teacher had given to solve this problem and I plugged it into a Photomath calculator, and the first step simplified the expression to ∛27/∛5. The eventual final answer was 3∛25/5, which is the correct answer listed in the back of the book, but I'm extremely confused as to how to get ∛27/∛5 from ∛54/∛10. Can anyone explain that first step to me? Thanks a bunch in advance.

Because I want to grind my axe on one point, I'll show you more than that.

Keep in mind that: \(\displaystyle \dfrac{54}{10} = \dfrac{27\cdot 2}{5\cdot 2} =\dfrac{27}{5}\cdot\dfrac{2}{2} = \dfrac{27}{5}\cdot 1 = \dfrac{27}{5}\). We'll need that act of changing the fraction to "Lowest Terms".

\(\displaystyle \dfrac{\sqrt[3]{54}}{\sqrt[3]{10}}= \sqrt[3]{\dfrac{54}{10}} = \sqrt[3]{\dfrac{27}{5}} = \dfrac{\sqrt[3]{27}}{\sqrt[3]{5}} =\dfrac{3}{\sqrt[3]{5}}\)


Personally, I think you are done at this point. However, there is an unnecessary requirement to Rationalize the Denominator. That exercise is where the final answer comes from.

 

[lolily]

Because I want to grind my axe on one point, I'll show you more than that.

Keep in mind that: \(\displaystyle \dfrac{54}{10} = \dfrac{27\cdot 2}{5\cdot 2} =\dfrac{27}{5}\cdot\dfrac{2}{2} = \dfrac{27}{5}\cdot 1 = \dfrac{27}{5}\). We'll need that act of changing the fraction to "Lowest Terms".

\(\displaystyle \dfrac{\sqrt[3]{54}}{\sqrt[3]{10}}= \sqrt[3]{\dfrac{54}{10}} = \sqrt[3]{\dfrac{27}{5}} = \dfrac{\sqrt[3]{27}}{\sqrt[3]{5}} =\dfrac{3}{\sqrt[3]{5}}\)


Personally, I think you are done at this point. However, there is an unnecessary requirement to Rationalize the Denominator. That exercise is where the final answer comes from.

Thank you so much, that makes perfect sense.