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Question Number 195799 by sonukgindia last updated on 10/Aug/23

Answered by som(math1967) last updated on 11/Aug/23

$$\frac{1}{a+\frac{1}{a}}+\frac{1}{a-\frac{1}{a}}=2a$$$$\Rightarrow \frac{a}{a^2+1}+\frac{a}{a^2-1}=2a$$$$\Rightarrow a[\frac{1}{a^2+1}+\frac{1}{a^2-1}]=2a$$$$\Rightarrow 2(a^4-1)=2a^2[takinga\ne 0]$$$$\Rightarrow a^4=a^2+1$$$$now{}^3\sqrt{\frac{1}{a^2+a+1}+\frac{1}{a^2-a+1}}$$$$={}^3\sqrt{\frac{2(a^2+1)}{(a^2+a+1)(a^2-a+1)}}$$$$={}^3\sqrt{\frac{2(a^2+1)}{a^4+a^2+1}}★$$$$={}^3\sqrt{\frac{2(a^2+1)}{a^2+1+a^2+1}}={}^3\sqrt{\frac{2(a^2+1)}{2(a^2+1)}}◼$$$$=1$$$$★(a^2+a+1)(a^2-a+1)$$$$={(a^2+1)}^2-a^2=a^4+a^2+1$$$$◼{\mathit{a}}^4={\mathit{a}}^2+1$$

Answered by Rasheed.Sindhi last updated on 13/Aug/23

$$\frac{1}{a+\frac{1}{a}}+\frac{1}{a-\frac{1}{a}}=2a$$$$\sqrt[3]{\frac{1}{a^2+a+1}+\frac{1}{a^2-a+1}}=?$$$$a-\frac{1}{a}+a+\frac{1}{a}=2a(a-\frac{1}{a})(a+\frac{1}{a})$$$$a^2-\frac{1}{a^2}=1\Rightarrow \frac{1}{a^2}=a^2-1$$$$\overline{)\begin{array}{c}\frac{1}{a^2}=a^2-1\end{array}}$$$$▸\sqrt[3]{\frac{1}{a^2+a+1}+\frac{1}{a^2-a+1}}$$$$=\sqrt[3]{\frac{1}{a(a+\frac{1}{a}+1)}+\frac{1}{a(a+\frac{1}{a}-1)}}$$$$=\sqrt[3]{\frac{a+\frac{1}{a}-1+a+\frac{1}{a}+1}{a(a+\frac{1}{a}+1)(a+\frac{1}{a}-1)}}$$$$=\sqrt[3]{\frac{2(a+\frac{1}{a})}{a({(a+\frac{1}{a})}^2-1)}}$$$$=\sqrt[3]{\frac{2(a+\frac{1}{a})}{a(a^2+\frac{1}{a^2}+1)}}$$$$=\sqrt[3]{\frac{2(a+\frac{1}{a})}{a(a^2+a^2-1+1)}}$$$$=\sqrt[3]{\frac{2(a+\frac{1}{a})}{2a^3}}$$$$=\sqrt[3]{\frac{a+\frac{1}{a}}{a^3}}=\sqrt[3]{\frac{a^2+1}{a^4}}=$$$$\sqrt[3]{\frac{1}{a^2}+{\left(\frac{1}{a^2}\right)}^2}$$$$=\sqrt[3]{a^2-1+{(a^2-1)}^2}$$$$=\sqrt[3]{a^2-1+a^4-2a^2+1}$$$$=\sqrt[3]{a^2(a^2-1)}=1$$$$=\sqrt[3]{a^2\left(\frac{1}{a^2}\right)}=1$$

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