2022/079) Simplify $(\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)} +1)^{48}$

 as we see that the roots are doubleing in term to term so multiply numeraator and denominator by $(\sqrt[16]{5}-1)$ we get

 $(\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)} +1)^{48}$

= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)(\sqrt[16]{5}-1)}  +1)^{48}$

= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[8]{5}-1)}  +1)^{48}$ using $a^2-b^2$ formula for last 2 terms

= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[4]{5}-1)}  +1)^{48}$ using $a^2-b^2$ formula for last 2 terms-

= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}  +1)^{48}$ using $a^2-b^2$ formula for last 2 terms-

= $(\frac{(4*\sqrt[16]{5}-1)}{4}+1)^{48}$ 

= $(\sqrt[16]{5})^{48} = 5^3 = 125$