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Question Number 191553 by MATHEMATICSAM last updated on 25/Apr/23
If a^2 + a + 1 = 0 then find a^5 + a^4 + 1.
$$\mathrm{If}\:{a}^{\mathrm{2}} \:+\:{a}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{find}\:{a}^{\mathrm{5}} \:+\:{a}^{\mathrm{4}} \:+\:\mathrm{1}. \\ $$
Commented by Tinku Tara last updated on 25/Apr/23
a=ω, ω^2 , where ω is cube root of unity.
$${a}=\omega,\:\omega^{\mathrm{2}} ,\:\mathrm{where}\:\omega\:\mathrm{is}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}. \\ $$
Answered by manxsol last updated on 27/Apr/23
(a−1)(a^2 +a+1)=a^3 −1=0 a^3 =1⇒a^2 (a_(=1) ^3 )+a(a_(=1) ^3 )+1=0
$$\left({a}−\mathrm{1}\right)\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)={a}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$${a}^{\mathrm{3}} =\mathrm{1}\Rightarrow{a}^{\mathrm{2}} \left({a}_{=\mathrm{1}} ^{\mathrm{3}} \right)+{a}\left({a}_{=\mathrm{1}} ^{\mathrm{3}} \right)+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$
Answered by mr W last updated on 25/Apr/23
a^2 +a+1=0 a^2 =−(a+1) a^4 =a^2 +2a+1=−a−1+2a+1=a a^5 =a^4 ×a=a^2 =−a−1 a^5 +a^4 +1=−a−1+a+1=0 ✓
$${a}^{\mathrm{2}} +{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} =−\left({a}+\mathrm{1}\right) \\ $$$${a}^{\mathrm{4}} ={a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}=−{a}−\mathrm{1}+\mathrm{2}{a}+\mathrm{1}={a} \\ $$$${a}^{\mathrm{5}} ={a}^{\mathrm{4}} ×{a}={a}^{\mathrm{2}} =−{a}−\mathrm{1} \\ $$$${a}^{\mathrm{5}} +{a}^{\mathrm{4}} +\mathrm{1}=−{a}−\mathrm{1}+{a}+\mathrm{1}=\mathrm{0}\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 26/Apr/23
Another way... a^2 + a + 1 = 0⇒1=−a^2 −a ▶a^5 + a^4 + 1=a^5 + a^4 −a^2 −a =a^3 (a^2 +a−(1/a) −(1/a^2 )) =a^3 {(a−(1/a))(a+(1/a))+(a−(1/a))} =a^3 (a−(1/a))(a+(1/a)+1) =a^3 (a−(1/a))(((a^2 +a+1)/a)) =a^2 (a−(1/a))(0)=0
$$\mathrm{Another}\:\mathrm{way}… \\ $$$$\:{a}^{\mathrm{2}} \:+\:{a}\:+\:\mathrm{1}\:=\:\mathrm{0}\Rightarrow\mathrm{1}=−{a}^{\mathrm{2}} −{a} \\ $$$$\blacktriangleright{a}^{\mathrm{5}} \:+\:{a}^{\mathrm{4}} \:+\:\mathrm{1}={a}^{\mathrm{5}} \:+\:{a}^{\mathrm{4}} −{a}^{\mathrm{2}} −{a} \\ $$$$\:\:\:={a}^{\mathrm{3}} \left({a}^{\mathrm{2}} +{a}−\frac{\mathrm{1}}{{a}}\:−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right) \\ $$$$\:\:\:={a}^{\mathrm{3}} \left\{\left({a}−\frac{\mathrm{1}}{{a}}\right)\left({a}+\frac{\mathrm{1}}{{a}}\right)+\left({a}−\frac{\mathrm{1}}{{a}}\right)\right\} \\ $$$$\:\:\:={a}^{\mathrm{3}} \left({a}−\frac{\mathrm{1}}{{a}}\right)\left({a}+\frac{\mathrm{1}}{{a}}+\mathrm{1}\right) \\ $$$$\:\:\:\:={a}^{\mathrm{3}} \left({a}−\frac{\mathrm{1}}{{a}}\right)\left(\frac{{a}^{\mathrm{2}} +{a}+\mathrm{1}}{{a}}\right) \\ $$$$\:\:\:\:={a}^{\mathrm{2}} \left({a}−\frac{\mathrm{1}}{{a}}\right)\left(\mathrm{0}\right)=\mathrm{0} \\ $$
Answered by Tinku Tara last updated on 05/Jul/23
a=w or w^2 a^5 +a^4 +1=w^2 +w+1=0
$${a}={w}\:{or}\:{w}^{\mathrm{2}} \\ $$$${a}^{\mathrm{5}} +{a}^{\mathrm{4}} +\mathrm{1}={w}^{\mathrm{2}} +{w}+\mathrm{1}=\mathrm{0} \\ $$

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