To the 48th power!

Algebra Level pending

Given that x = 4 ( 5 + 1 ) ( 5 4 + 1 ) ( 5 8 + 1 ) ( 5 16 + 1 ) x= \frac { 4 }{ (\sqrt { 5 } +1)(\sqrt [ 4 ]{ 5 } +1)(\sqrt [ 8 ]{ 5 } +1)(\sqrt [ 16 ]{ 5 } +1) }

Find ( x + 1 ) 48 (x+1)^{48}


The answer is 125.

William Isoroku by William Isoroku , 25, USA

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2 solutions

Alex Delhumeau
May 12, 2015

4 ( 5 + 1 ) ( 5 4 + 1 ) ( 5 8 + 1 ) ( 5 16 + 1 ) = \large{\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}}=

4 ( 5 − 1 ) ( 5 4 − 1 ) ( 5 8 − 1 ) ( 5 16 − 1 ) ( 5 + 1 ) ( 5 − 1 ) ( 5 4 + 1 ) ( 5 4 − 1 ) ( 5 8 + 1 ) ( 5 8 − 1 ) ( 5 16 + 1 ) ( 5 16 − 1 ) = \large{4\frac{(\sqrt{5}-1)(\sqrt[4]{5}-1)(\sqrt[8]{5}-1)(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)(\sqrt[4]{5}+1)(\sqrt[4]{5}-1)(\sqrt[8]{5}+1)(\sqrt[8]{5}-1)(\sqrt[16]{5}+1)(\sqrt[16]{5}-1)}}=

4 ( 5 − 1 ) ( 5 4 − 1 ) ( 5 8 − 1 ) ( 5 16 − 1 ) ( 5 − 1 ) ( 5 − 1 ) ( 5 4 − 1 ) ( 5 8 − 1 ) = \large{4\frac{(\sqrt{5}-1)(\sqrt[4]{5}-1)(\sqrt[8]{5}-1)(\sqrt[16]{5}-1)}{(5-1)(\sqrt{5}-1)(\sqrt[4]{5}-1)(\sqrt[8]{5}-1)}}=

5 16 − 1 → x = 5 16 \sqrt[16]{5}-1 \rightarrow x = \sqrt[16]{5} .

5 16 48 = 5 3 = 125 {\sqrt[16]{5}}^{48}=5^3=\boxed{\small{125}}

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L e t t = 5 16 ∴ X = 4 ( t 8 + 1 ) ( t 4 + 1 ) ( t 2 + 1 ) ( t + 1 ) Multiplying numerator and denominator by ( t − 1 ) ⟹ Let~ t=\sqrt[16]5~~~~~\therefore ~X=\large \dfrac{4}{(t^8+1)(t^4+1)(t^2+1)(t+1)} \\\text{Multiplying numerator and denominator by } (t-1)\\\implies

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