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$$x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}.$$

Find $$(x+1)^{48}$$

Jan 15, 2019
edited by vindou  Jan 15, 2019

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In the first equation, multiply by $$-\sqrt[16]{5}+1$$ to cancel out all of the nasty roots and you get $$\sqrt[16]{5}x + x = 1$$. Simplifying further results in $$x = \sqrt[16]{5}-1$$ .

Since we want $$(x+1)^{48}$$, we have $$(\sqrt[16]{5})^{48}$$, which equals $$5^3 = \boxed{125}$$

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Jan 15, 2019
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Thank you so much!

vindou  Jan 15, 2019