If a^2 + a + 1 = 0 then find a^5 + a^4 + 1.


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Question Number 191553 by MATHEMATICSAM last updated on 25/Apr/23

$If a^{2} + a + 1 = 0 then find a^{5} + a^{4} + 1 .$
Commented by Tinku Tara last updated on 25/Apr/23

$a = ω, ω^{2}, where ω is cube root of unity .$
	Answered by manxsol last updated on 27/Apr/23

	$(a - 1) (a^{2} + a + 1) = a^{3} - 1 = 0$ $a^{3} = 1 \Rightarrow a^{2} (a_{= 1}^{3}) + a (a_{= 1}^{3}) + 1 = 0$
	Answered by mr W last updated on 25/Apr/23

	$a^{2} + a + 1 = 0$ $a^{2} = - (a + 1)$ $a^{4} = a^{2} + 2 a + 1 = - a - 1 + 2 a + 1 = a$ $a^{5} = a^{4} \times a = a^{2} = - a - 1$ $a^{5} + a^{4} + 1 = - a - 1 + a + 1 = 0 ✓$
	Answered by Rasheed.Sindhi last updated on 26/Apr/23
	$Another way... a^2 + a + 1 = 0⇒1=−a^2 −a ▶a^5 + a^4 + 1=a^5 + a^4 −a^2 −a =a^3 (a^2 +a−(1/a) −(1/a^2 )) =a^3 {(a−(1/a))(a+(1/a))+(a−(1/a))} =a^3 (a−(1/a))(a+(1/a)+1) =a^3 (a−(1/a))(((a^2 +a+1)/a)) =a^2 (a−(1/a))(0)=0$
	$Another way . . .$ $a^{2} + a + 1 = 0 \Rightarrow 1 = - a^{2} - a$ $▸ a^{5} + a^{4} + 1 = a^{5} + a^{4} - a^{2} - a$ $= a^{3} (a^{2} + a - \frac{1}{a} - \frac{1}{a^{2}})$ $= a^{3} {(a - \frac{1}{a}) (a + \frac{1}{a}) + (a - \frac{1}{a})}$ $= a^{3} (a - \frac{1}{a}) (a + \frac{1}{a} + 1)$ $= a^{3} (a - \frac{1}{a}) (\frac{a^{2} + a + 1}{a})$ $= a^{2} (a - \frac{1}{a}) (0) = 0$
	Answered by Tinku Tara last updated on 05/Jul/23

	$a = w o r w^{2}$ $a^{5} + a^{4} + 1 = w^{2} + w + 1 = 0$
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