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Question Number 186105 by HeferH last updated on 01/Feb/23
a^2 + 1 = a a^(12) + a^6 + 1 = ?
$$\:{a}^{\mathrm{2}} \:+\:\mathrm{1}\:\:=\:{a} \\ $$$$\:{a}^{\mathrm{12}} \:+\:{a}^{\mathrm{6}} \:+\:\mathrm{1}\:=\:?\: \\ $$
Answered by Rasheed.Sindhi last updated on 01/Feb/23
a^2 +1=a; a^(12) +a^6 +1=? a^2 =a−1 a^3 =a^2 −a=(a−1)−a=−1 ▶a^(12) +a^6 +1=(a^3 )^4 +(a^3 )^2 +1 =(−1)^4 +(−1)^2 +1=3
$${a}^{\mathrm{2}} +\mathrm{1}={a};\:{a}^{\mathrm{12}} +{a}^{\mathrm{6}} +\mathrm{1}=? \\ $$$$\: \\ $$$${a}^{\mathrm{2}} ={a}−\mathrm{1} \\ $$$${a}^{\mathrm{3}} ={a}^{\mathrm{2}} −{a}=\left({a}−\mathrm{1}\right)−{a}=−\mathrm{1} \\ $$$$\: \\ $$$$\blacktriangleright{a}^{\mathrm{12}} +{a}^{\mathrm{6}} +\mathrm{1}=\left({a}^{\mathrm{3}} \right)^{\mathrm{4}} +\left({a}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{1} \\ $$$$\:\:\:\:=\left(−\mathrm{1}\right)^{\mathrm{4}} +\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{3} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Feb/23
a^2 −a+1=0⇒a+(1/a)=1 ; a≠0 (a+(1/a))^2 =1^2 ⇒a^2 +(1/a^2 )=−1 (a^2 +(1/a^2 ))^3 =(−1)^3 a^6 +(1/a^6 )+3(a^2 +(1/a^2 ))=−1 a^6 +(1/a^6 )+3(−1)=−1⇒a^6 +(1/a^6 )=2 a^(12) −2a^6 +1=0 (a^6 −1)^2 =0 a^6 =1 a^(12) +a^6 +1=a^6 (a^6 +(1/a^6 )+1) =(1)(2+1)=3
$${a}^{\mathrm{2}} −{a}+\mathrm{1}=\mathrm{0}\Rightarrow{a}+\frac{\mathrm{1}}{{a}}=\mathrm{1}\:;\:{a}\neq\mathrm{0} \\ $$$$\:\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=−\mathrm{1} \\ $$$$\left({a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)^{\mathrm{3}} =\left(−\mathrm{1}\right)^{\mathrm{3}} \\ $$$${a}^{\mathrm{6}} +\frac{\mathrm{1}}{{a}^{\mathrm{6}} }+\mathrm{3}\left({a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)=−\mathrm{1} \\ $$$${a}^{\mathrm{6}} +\frac{\mathrm{1}}{{a}^{\mathrm{6}} }+\mathrm{3}\left(−\mathrm{1}\right)=−\mathrm{1}\Rightarrow{a}^{\mathrm{6}} +\frac{\mathrm{1}}{{a}^{\mathrm{6}} }=\mathrm{2} \\ $$$${a}^{\mathrm{12}} −\mathrm{2}{a}^{\mathrm{6}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({a}^{\mathrm{6}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{6}} =\mathrm{1} \\ $$$$ \\ $$$${a}^{\mathrm{12}} +{a}^{\mathrm{6}} +\mathrm{1}={a}^{\mathrm{6}} \left({a}^{\mathrm{6}} +\frac{\mathrm{1}}{{a}^{\mathrm{6}} }+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}\right)\left(\mathrm{2}+\mathrm{1}\right)=\mathrm{3} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Feb/23
a^2 + 1 = a; a^(12) + a^6 + 1 = ? (a^2 +1)^2 =a^2 a^4 +2a^2 +1=a^2 a^4 +1=−a^2 (a^4 +1)^3 =(−a^2 )^3 a^(12) +1+3a^4 (a^4 +1)=−a^6 a^(12) +1+3a^4 (−a^2 )=−a^6 a^(12) −2a^6 +1=0 (a^6 −1)^2 =0 a^6 =1 ▶a^(12) +a^6 +1=(a^6 )^2 +a^6 +1 =(1)^2 +1+1=3
$$\:{a}^{\mathrm{2}} \:+\:\mathrm{1}\:\:=\:{a};\:\:\:\:{a}^{\mathrm{12}} \:+\:{a}^{\mathrm{6}} \:+\:\mathrm{1}\:=\:? \\ $$$$\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}={a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{4}} +\mathrm{1}=−{a}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}} =\left(−{a}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$${a}^{\mathrm{12}} +\mathrm{1}+\mathrm{3}{a}^{\mathrm{4}} \left({a}^{\mathrm{4}} +\mathrm{1}\right)=−{a}^{\mathrm{6}} \\ $$$${a}^{\mathrm{12}} +\mathrm{1}+\mathrm{3}{a}^{\mathrm{4}} \left(−{a}^{\mathrm{2}} \right)=−{a}^{\mathrm{6}} \\ $$$${a}^{\mathrm{12}} −\mathrm{2}{a}^{\mathrm{6}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({a}^{\mathrm{6}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{6}} =\mathrm{1} \\ $$$$\blacktriangleright{a}^{\mathrm{12}} +{a}^{\mathrm{6}} +\mathrm{1}=\left({a}^{\mathrm{6}} \right)^{\mathrm{2}} +{a}^{\mathrm{6}} +\mathrm{1} \\ $$$$=\left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}+\mathrm{1}=\mathrm{3} \\ $$
Commented by HeferH last updated on 01/Feb/23
(thank you)^3 !
$$\left({thank}\:{you}\right)^{\mathrm{3}} \:! \\ $$

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