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Question-195799

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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
(1/(a+(1/a))) +(1/(a−(1/a)))=2a ⇒ (a/(a^2 +1)) +(a/(a^2 −1))=2a ⇒a[(1/(a^2 +1)) +(1/(a^2 −1))]=2a ⇒2(a^4 −1)=2a^2 [taking a≠0] ⇒a^4 =a^2 +1 now ^3 (√((1/(a^2 +a+1))+(1/(a^2 −a+1)))) =^3 (√((2(a^2 +1))/((a^2 +a+1)(a^2 −a+1)))) =^3 (√((2(a^2 +1))/(a^4 +a^2 +1))) ★ =^3 (√((2(a^2 +1))/(a^2 +1+a^2 +1)))=^3 (√((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 ■ a^4 =a^2 +1
$$\:\frac{\mathrm{1}}{{a}+\frac{\mathrm{1}}{{a}}}\:+\frac{\mathrm{1}}{{a}−\frac{\mathrm{1}}{{a}}}=\mathrm{2}{a} \\ $$$$\Rightarrow\:\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2}{a} \\ $$$$\Rightarrow{a}\left[\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\right]=\mathrm{2}{a} \\ $$$$\Rightarrow\mathrm{2}\left({a}^{\mathrm{4}} −\mathrm{1}\right)=\mathrm{2}{a}^{\mathrm{2}} \:\:\:\left[{taking}\:{a}\neq\mathrm{0}\right] \\ $$$$\Rightarrow{a}^{\mathrm{4}} ={a}^{\mathrm{2}} +\mathrm{1} \\ $$$${now}\:\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{a}+\mathrm{1}}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{a}+\mathrm{1}}} \\ $$$$=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}{\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} −{a}+\mathrm{1}\right)}} \\ $$$$=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}{{a}^{\mathrm{4}} +{a}^{\mathrm{2}} +\mathrm{1}}}\:\:\:\:\bigstar \\ $$$$=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}{{a}^{\mathrm{2}} +\mathrm{1}+{a}^{\mathrm{2}} +\mathrm{1}}}=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}}\:\blacksquare \\ $$$$=\:\mathrm{1} \\ $$$$\bigstar\:\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} −{a}+\mathrm{1}\right) \\ $$$$=\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} ={a}^{\mathrm{4}} +{a}^{\mathrm{2}} +\mathrm{1} \\ $$$$\blacksquare\:\boldsymbol{{a}}^{\mathrm{4}} =\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Aug/23
(1/(a+(1/a)))+(1/(a−(1/a)))=2a (((1/(a^2 +a+1))+(1/(a^2 −a+1))))^(1/3) =? a−(1/a)+a+(1/a)=2a(a−(1/a))(a+(1/a)) a^2 −(1/a^2 )=1⇒(1/a^2 )=a^2 −1 determinant ((((1/a^2 )=a^2 −1))) ▶(((1/(a^2 +a+1))+(1/(a^2 −a+1))))^(1/3) =(((1/(a(a+(1/a)+1)))+(1/(a(a+(1/a)−1)))))^(1/3) =(((a+(1/a)−1+a+(1/a)+1)/(a(a+(1/a)+1)(a+(1/a)−1))))^(1/3) =(((2(a+(1/a)))/(a((a+(1/a))^2 −1))))^(1/3) =(((2(a+(1/a)))/(a(a^2 +(1/a^2 )+1))))^(1/3) =(((2(a+(1/a)))/(a(a^2 +a^2 −1+1))))^(1/3) =(((2(a+(1/a)))/(2a^3 )))^(1/3) =(((a+(1/a))/a^3 ))^(1/3) =(((a^2 +1)/a^4 ))^(1/3) = (((1/a^2 )+((1/a^2 ))^2 ))^(1/3) =((a^2 −1+(a^2 −1)^2 ))^(1/3) =((a^2 −1+a^4 −2a^2 +1))^(1/3) =((a^2 (a^2 −1)))^(1/3) =1 =((a^2 ((1/a^2 ))))^(1/3) =1
$$\frac{\mathrm{1}}{{a}+\frac{\mathrm{1}}{{a}}}+\frac{\mathrm{1}}{{a}−\frac{\mathrm{1}}{{a}}}=\mathrm{2}{a} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{a}+\mathrm{1}}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{a}+\mathrm{1}}}\:=? \\ $$$${a}−\frac{\mathrm{1}}{{a}}+{a}+\frac{\mathrm{1}}{{a}}=\mathrm{2}{a}\left({a}−\frac{\mathrm{1}}{{a}}\right)\left({a}+\frac{\mathrm{1}}{{a}}\right) \\ $$$${a}^{\mathrm{2}} −\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\mathrm{1}\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }={a}^{\mathrm{2}} −\mathrm{1} \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }={a}^{\mathrm{2}} −\mathrm{1}}\\\hline\end{array} \\ $$$$\blacktriangleright\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{a}+\mathrm{1}}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{a}+\mathrm{1}}}\: \\ $$$$\:\:=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{a}\left({a}+\frac{\mathrm{1}}{{a}}+\mathrm{1}\right)}+\frac{\mathrm{1}}{{a}\left({a}+\frac{\mathrm{1}}{{a}}−\mathrm{1}\right)}}\: \\ $$$$\:\:=\sqrt[{\mathrm{3}}]{\frac{{a}+\frac{\mathrm{1}}{{a}}−\mathrm{1}+{a}+\frac{\mathrm{1}}{{a}}+\mathrm{1}}{{a}\left({a}+\frac{\mathrm{1}}{{a}}+\mathrm{1}\right)\left({a}+\frac{\mathrm{1}}{{a}}−\mathrm{1}\right)}}\: \\ $$$$\:\:=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\left({a}+\frac{\mathrm{1}}{{a}}\right)}{{a}\left(\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} −\mathrm{1}\right)}}\: \\ $$$$\:\:=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\left({a}+\frac{\mathrm{1}}{{a}}\right)}{{a}\left({a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\mathrm{1}\right)}}\: \\ $$$$\:\:=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\left({a}+\frac{\mathrm{1}}{{a}}\right)}{{a}\left({a}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right)}}\: \\ $$$$\:\:=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\left({a}+\frac{\mathrm{1}}{{a}}\right)}{\mathrm{2}{a}^{\mathrm{3}} }}\: \\ $$$$\:\:=\sqrt[{\mathrm{3}}]{\frac{{a}+\frac{\mathrm{1}}{{a}}}{{a}^{\mathrm{3}} }}\:=\sqrt[{\mathrm{3}}]{\frac{{a}^{\mathrm{2}} +\mathrm{1}}{{a}^{\mathrm{4}} }}\:= \\ $$$$\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} }\: \\ $$$$=\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} −\mathrm{1}+\left({a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} −\mathrm{1}+{a}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} −\mathrm{1}\right)}\:\:=\mathrm{1} \\ $$$$=\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)}\:\:=\mathrm{1} \\ $$

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