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Simplify $$\frac{1}{\sqrt[3]{2}+\sqrt[3]{16}}$$.

 Aug 12, 2023

Best Answer 

 #4
avatar+129197 
+1

cube root 16 =  2cube root 2

 

So the expression simplifies to    1 / [3 * 3(2)]......multiply top/bottom by  3(2^2) and we  have

 

3(2^2) / 3 * 3(2^3)  = 

 

3(2^2) / 3* 2  =

 

3 [ 4 ] / 6

 

 

cool cool cool

 Aug 13, 2023
 #1
avatar
+1

 

We can write \begin{align*} \frac{1}{\sqrt[3]{2}+\sqrt[3]{16}} &= \frac{1}{\sqrt[3]{2}+\sqrt[3]{4^2}}\ &= \frac{1}{\sqrt[3]{2}+2}\ &= \frac{\sqrt[3]{2}-2}{2(\sqrt[3]{2}-2)}\ &= \boxed{\frac{1}{2}}. \end{align*}

 Aug 12, 2023
 #3
avatar+757 
0

That answer is wrong because 3sqrt4^2 is not equal to 2. 

history  Aug 13, 2023
 #4
avatar+129197 
+1
Best Answer

cube root 16 =  2cube root 2

 

So the expression simplifies to    1 / [3 * 3(2)]......multiply top/bottom by  3(2^2) and we  have

 

3(2^2) / 3 * 3(2^3)  = 

 

3(2^2) / 3* 2  =

 

3 [ 4 ] / 6

 

 

cool cool cool

CPhill Aug 13, 2023

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