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Question Number 146896 by mathdanisur last updated on 16/Jul/21

$$\frac{5}{\sqrt[8]{6}+1}\cdot \frac{1}{\sqrt[4]{6}+1}\cdot \frac{1}{\sqrt{6}+1}+1=?$$

Answered by EDWIN88 last updated on 16/Jul/21

$$\mathrm{What}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\frac{5}{\sqrt[8]{6}+1}\times \frac{1}{\sqrt[4]{6}+1}\times \frac{1}{\sqrt{6}+1}+1.$$$$\mathrm{Solution}:$$$$\mathrm{let}\sqrt[8]{6}=\mathrm{u}\to \{\begin{array}{l}\sqrt[4]{6}={\mathrm{u}}^2\\ \sqrt{6}={\mathrm{u}}^4\end{array}$$$$\mathrm{T}=\frac{5}{\sqrt[8]{6}+1}\times \frac{1}{\sqrt[4]{6}+1}\times \frac{1}{\sqrt{6}+1}+1$$$$\mathrm{T}=\frac{5}{\mathrm{u}+1}\times \frac{1}{{\mathrm{u}}^2+1}\times \frac{1}{{\mathrm{u}}^4+1}+1$$$$\mathrm{T}=\frac{5}{(\mathrm{u}+1)({\mathrm{u}}^2+1)({\mathrm{u}}^4+1)}+1$$$$\mathrm{T}=\frac{5}{(\mathrm{u}+1)({\mathrm{u}}^6+{\mathrm{u}}^4+{\mathrm{u}}^2+1)}+1$$$$\mathrm{T}=\frac{5}{{\mathrm{u}}^7+{\mathrm{u}}^6+{\mathrm{u}}^5+{\mathrm{u}}^4+{\mathrm{u}}^3+{\mathrm{u}}^2+\mathrm{u}+1}+1$$$$\mathrm{T}=\frac{5(\mathrm{u}-1)}{{\mathrm{u}}^8-1}+1$$$$\mathrm{T}=\frac{5(\sqrt[8]{6}-1)}{6-1}+1=\sqrt[8]{6}$$

Commented by mathdanisur last updated on 16/Jul/21

$$thankyouSer$$

Answered by liberty last updated on 16/Jul/21

$$\frac{5}{\sqrt[8]{6}+1}\times \frac{\sqrt[8]{6}-1}{\sqrt[8]{6}-1}\times \frac{1}{\sqrt[4]{6}+1}\times \frac{1}{\sqrt{6}+1}+1$$$$=\frac{5(\sqrt[8]{6}-1)}{\sqrt[4]{6}-1}\times \frac{1}{\sqrt[4]{6}+1}\times \frac{1}{\sqrt{6}+1}+1$$$$=\frac{5(\sqrt[8]{6}-1)}{\sqrt{6}-1}\times \frac{1}{\sqrt{6}+1}+1$$$$=\frac{5(\sqrt[8]{6}-1)}{6-1}+1$$$$=\sqrt[8]{6}-1+1=\sqrt[8]{6}.$$

Commented by mathdanisur last updated on 16/Jul/21

$$thankyouSer$$

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