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Let\[\begin{aligned} a &= \sqrt{2}+\sqrt{3}+\sqrt{6}, \\ b &= -\sqrt{2}+\sqrt{3}+\sqrt{6}, \\ c&= \sqrt{2}-\sqrt{3}+\sqrt{6}, \\ d&=-\sqrt{2}-\sqrt{3}+\sqrt{6}. \end{aligned}\]Evaluate $\left(\frac1a + \frac1b + \frac1c + \frac1d\right)^2.$

 May 21, 2022
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\(\begin{align*} \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} &= \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{6}} + \frac{1}{-\sqrt{2} + \sqrt{3} + \sqrt{6}} + \frac{1}{\sqrt{2} - \sqrt{3} + \sqrt{6}} + \frac{1}{-\sqrt{2} - \sqrt{3} + \sqrt{6}} \\ &= \frac{-12 + 7 \sqrt{2} + 5 \sqrt{3} - \sqrt{6}}{23} + \frac{12 - 7 \sqrt{2} + 5 \sqrt{3} - \sqrt{6}}{23} + \frac{12 + 5 \sqrt{2} - 7 \sqrt{3} - 2 \sqrt{6}}{23} + \frac{-12 - 5 \sqrt{2} + 7 \sqrt{3} - 2 \sqrt{6}}{23} \\ &= \frac{-6 \sqrt{6}}{23}. \end{align*}\)

Therefore,
\(\left( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} \right)^2 = \dfrac{216}{529}.\)

 May 21, 2022

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