Radical Expression Simplification 4
If the number \[\frac{1}{2} \left(\frac{5}{\sqrt[3]{3} + \sqrt[3]{2}} + \frac1{\sqrt[3]{3} -\sqrt[3]{2}}\right)\]can be expressed in the form $ \sqrt[3]{a} + \sqrt[3]{b}, $ where $ a $ and $ b $ are integers, compute $ a+b $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
We rationalize each of the fractions in parentheses, by using the sum and difference of cubes identities. First, we have \[\begin{aligned} \frac{5}{\sqrt[3]{3} + \sqrt[3]{2}} &= \frac{5\left(\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}\right)}{\left(\sqrt[3]{3} + \sqrt[3]{2}\right)\left(\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}\right)} \\ &= \frac{5\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}{3+2} \\ &= \sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}.\end{aligned}\]Similarly, \[\begin{aligned} \frac{1}{\sqrt[3]{3} - \sqrt[3]{2}} &= \frac{\sqrt[3]{9} + \sqrt[3]{6} + \sqrt[3]{4}}{\left(\sqrt[3]{3} - \sqrt[3]{2}\right)\left(\sqrt[3]{9} + \sqrt[3]{6} + \sqrt[3]{4}\right)} \\ &= \frac{\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}}{3 - 2} \\ &= \sqrt[3]{9} + \sqrt[3]{6} + \sqrt[3]{4}.\end{aligned}\]Therefore,\[\begin{aligned} \frac{1}{2} \left(\frac{5}{\sqrt[3]{3} + \sqrt[3]{2}} + \frac1{\sqrt[3]{3} -\sqrt[3]{2}}\right) &= \frac{1}{2} \left(\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right) + \left(\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}\right) \right) \\ &= \sqrt[3]{9} + \sqrt[3]{4}, \end{aligned}\]so $ a+b=9+4=\boxed{13} $.
Freeflows