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Question Number 126657 by adeyemiprof40 last updated on 23/Dec/20

Answered by Olaf last updated on 23/Dec/20

a^4 +a^3 +a^2 +a+1 = 0 ⇔ ((a^5 −1)/(a−1)) = 0, a≠1 ⇔ a^5 = 1 a≠1 a = e^((2/5)ikπ) , k = 1, 2, 3, 4 a^(2.....000) +a^(20....010) +1 = 1+1+1 = 3

$${a}^{\mathrm{4}} +{a}^{\mathrm{3}} +{a}^{\mathrm{2}} +{a}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:\frac{{a}^{\mathrm{5}} −\mathrm{1}}{{a}−\mathrm{1}}\:=\:\mathrm{0},\:{a}\neq\mathrm{1} \\ $$$$\Leftrightarrow\:{a}^{\mathrm{5}} \:=\:\mathrm{1}\:\:{a}\neq\mathrm{1} \\ $$$${a}\:=\:{e}^{\frac{\mathrm{2}}{\mathrm{5}}{ik}\pi} ,\:{k}\:=\:\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4} \\ $$$${a}^{\mathrm{2}.....\mathrm{000}} +{a}^{\mathrm{20}....\mathrm{010}} +\mathrm{1}\:=\:\mathrm{1}+\mathrm{1}+\mathrm{1}\:=\:\mathrm{3} \\ $$

Answered by AlagaIbile last updated on 23/Dec/20

a^5 = 1 Therefore the require answer is ⇒ 1 + 1 + 1 = 3

$$\:{a}^{\mathrm{5}} \:=\:\mathrm{1} \\ $$$$\:{Therefore}\:{the}\:{require}\:{answer}\:{is} \\ $$$$\Rightarrow\:\mathrm{1}\:+\:\mathrm{1}\:+\:\mathrm{1}\:=\:\mathrm{3} \\ $$

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