Question

Q. Find the value of
8. $\frac{\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}}}{\sqrt[3]{8}}=?$

Answer 1

$$\begin{gathered} \frac{\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}}}{\sqrt[3]{8}} \\ =\frac{\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+15}}}}}{\sqrt[3]{2\times 2\times 2}}\mathbf{}\left[\mathbf{as}\mathbf{,}\mathbf{}\sqrt{\mathbf{225}}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{15}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{15}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{15}\right] \\ =\frac{{\displaystyle \sqrt{10+\sqrt{25+\sqrt{108+\sqrt{169}}}}}}{{\displaystyle 2}} \\ =\frac{{\displaystyle \sqrt{10+\sqrt{25+\sqrt{108+13}}}}}{{\displaystyle 2}}\mathbf{}\mathbf{}\mathbf{}\left[\mathbf{as}\mathbf{,}\mathbf{}\sqrt{\mathbf{169}}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{13}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{13}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{13}\right] \\ =\frac{{\displaystyle \sqrt{10+\sqrt{25+\sqrt{121}}}}}{{\displaystyle 2}} \\ =\frac{{\displaystyle \sqrt{10+\sqrt{25+11}}}}{{\displaystyle 2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\mathbf{as}\mathbf{,}\mathbf{}\sqrt{\mathbf{121}}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{11}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{11}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{11}\right] \\ =\frac{\sqrt{10+\sqrt{36}}}{2} \\ =\frac{\sqrt{10+6}}{2}\mathbf{}\mathbf{}\left[\mathbf{as}\mathbf{,}\mathbf{}\sqrt{\mathbf{36}}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{6}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{6}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}\right] \\ =\frac{\sqrt{16}}{2} \\ =\frac{\sqrt{4\times 4}}{2} \\ =\frac{4}{2} \\ =2 \end{gathered}$$