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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
 #1
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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 \(5^3 = \boxed{125}\)

 

-24

 Jan 15, 2019
 #2
avatar+62 
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Thank you so much!

vindou  Jan 15, 2019

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