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Rationalize the denominator of $\frac{1}{\sqrt[3]{2} + \sqrt{16}}$

 Nov 29, 2021
 #1
avatar+168 
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\({ \ {1} \over 3 root 2 +4}\\\)

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 Nov 29, 2021
 #2
avatar+69 
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Actualy rationalizing the denominator means getting rid of the roots, you did not do that.

 Nov 29, 2021
 #3
avatar+117437 
+1

The sum of 2 cubics is  ..  (something you need to memorize)

\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

 

So if you have a denominator of  \((a+b)\) where a or b have cube roots, and you multiply it by  \((a^2-ab+b^2)\) then you will end up with a rational denominator.

 

\(\frac{1}{\sqrt[3]{2}+\sqrt{16}}=\frac{1}{\sqrt[3]{2}+4}\)

 

let

\(a= 2^{1/3}\qquad    and\qquad b=4\\so\\ a^2=2^{2/3}\qquad    and\qquad b^2=16\\ ab=4*2^{1/3}\)

 

\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

 

\(\frac{1}{\sqrt[3]{2}+4}\times \frac{2^{2/3}-4*2^{(1/3)}+4^2}{2^{2/3}-4*2^{(1/3)}+4^2}\\~\\ =\frac{2^{2/3}-4*2^{(1/3)}+4^2}{2+64}\\~\\ =\frac{\sqrt[3]{4}-4\sqrt[3]{2}+16}{66}\\~\\\)

 

Here is a video that might help you understand

https://www.youtube.com/watch?v=33xwVWyrFgg

 Nov 29, 2021

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