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Question Number 146622 by mathdanisur last updated on 14/Jul/21

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

Commented by hknkrc46 last updated on 14/Jul/21

$$★\frac{1}{(\sqrt[8]{6}+1)(\sqrt[4]{6}+1)(\sqrt{6}+1)}+1$$$$=\frac{\sqrt[8]{6}-1}{\underset{\sqrt[4]{6}-1}{\underbrace{(\sqrt[8]{6}-1)(\sqrt[8]{6}+1)}}(\sqrt[4]{6}+1)(\sqrt{6}+1)}+1$$$$=\frac{\sqrt[8]{6}-1}{\underset{\sqrt{6}-1}{\underbrace{(\sqrt[4]{6}-1)(\sqrt[4]{6}+1)}}(\sqrt{6}+1)}+1$$$$=\frac{\sqrt[8]{6}-1}{(\sqrt{6}-1)(\sqrt{6}+1)}+1$$$$=\frac{\sqrt[8]{6}-1}{6-1}+1=\frac{\sqrt[8]{6}-1}{5}+1$$$$=\frac{\sqrt[8]{6}+4}{5}$$

Answered by som(math1967) last updated on 14/Jul/21

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

Commented by mathdanisur last updated on 14/Jul/21

$$thanksSer,butanswer:\sqrt[8]{6}$$

Commented by som(math1967) last updated on 14/Jul/21

$$whereis\mathit{w}\mathit{r}\mathit{o}\mathit{n}\mathit{g}?$$

Commented by mathdanisur last updated on 14/Jul/21

$$Ser,maybetheansweriswrong$$

Commented by som(math1967) last updated on 14/Jul/21

$$let\frac{1}{({}^8\sqrt{6}+1)({}^4\sqrt{6}+1)(\sqrt{6}+1)}+1{=}^8\sqrt{6}$$$$\frac{1}{({}^8\sqrt{6}+1)({}^4\sqrt{6}+1)(\sqrt{6}+1)}{=}^8\sqrt{6}-1$$$$\frac{1}{({}^8\sqrt{6}+1)({}^4\sqrt{6}+1)(\sqrt{6}+1)({}^8\sqrt{6}-1)}=1$$$$\frac{1}{6-1}=1$$$$\therefore \frac{1}{5}=1??????$$$$soIthinkansiswrong$$

Commented by mathdanisur last updated on 14/Jul/21

$$answerwrongSer,thankyou$$

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