Question

If $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{45}+1)(\sqrt[8]{85}+1)(\sqrt[16]{165}+1)}$. Then the value of $(1+x{)}^{48}$ is

• A. $5$
• B. $25$
• C. $125$
• D. $625$

Answer 1

Answer: (C)

$x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{45}+1)(\sqrt[8]{85}+1)(\sqrt[16]{165}+1)}$

$\Rightarrow x=\frac{4(\sqrt[16]{165}-1)}{(\sqrt{5}+1)(\sqrt[4]{45}+1)(\sqrt[8]{85}+1)(\sqrt[16]{165}+1)(\sqrt[16]{165}-1)}$

$=\frac{4(\sqrt[16]{165}-1)}{(\sqrt{5}+1)(\sqrt[4]{45}+1)(\sqrt[8]{85}+1)(\sqrt[8]{85}-1)}$

$=\frac{4(\sqrt[16]{165}-1)}{(\sqrt{5}+1)(\sqrt[4]{45}+1)(\sqrt[4]{45}-1)}$

$\Rightarrow x=\frac{4(\sqrt[16]{165}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}$

$=\frac{4(\sqrt[16]{165}-1)}{5-1}$

$=\frac{4(\sqrt[16]{165}-1)}{4}$

$\Rightarrow x+1=\sqrt[16]{165}-1+1=\sqrt[16]{165}$

So, ${(1+x)}^{48}$

$={\left(\sqrt[16]{165}\right)}^{48}={\left(5\right)}^{48/16}=5^3=125$

Hence, the answer is $125.$