Assuming that 'a' is real, a^5 = 1 implies that a = 1. but it had been shown that 'a' not equal 1. by inspection, the solution for real 'a' is approximately slightly less than -1 (somewhere between -2 and -1).
1+a+a^2+a^3+a^4=0 | recognize the general form (geometric series): sum_(k=0)^(n) a^k=1+a+a^2+...+a^n=(a^(n+1)-1)/(a-1) so multiply both sides: (a-1)(1+a+a^2+a^3+a^4)=0 thus: a^5-1=0 So a^5=1 Thus a^2020+a^2010+1= (a^5)^(404)+(a^5)^(402)+1= 1^404+1^402+1= 1+1+1= 3
Assuming that 'a' is real, a^5 = 1 implies that a = 1. but it had been shown that 'a' not equal 1. by inspection, the solution for real 'a' is approximately slightly less than -1 (somewhere between -2 and -1).
1+a+a^2+a^3+a^4=0 | recognize the general form (geometric series): sum_(k=0)^(n) a^k=1+a+a^2+...+a^n=(a^(n+1)-1)/(a-1) so multiply both sides:
(a-1)(1+a+a^2+a^3+a^4)=0
thus:
a^5-1=0
So a^5=1
Thus a^2020+a^2010+1=
(a^5)^(404)+(a^5)^(402)+1=
1^404+1^402+1=
1+1+1=
3
I have another solution
Action 1:
a^4 + a^3 + a^2 + a + 1 = 0
a^2 * (a+1) + a + 1 = -(a^4)
(a^2 + 1) * (a+1) = -(a^4).
Action 2:
a^4 + a^3 + a^2 + a + 1 = 0
a^2 * (a^2 + 1) + a * (a^2 + 1) = -1
a * (a+1) * (a^2 + 1) = -1
(a+1) * (a^2 + 1) = -(1/a).
And clearly 'a' can't equal zero. So we put Action 1 and Action 2 together...
Action 3:
a^4 = 1/a
a^5 = 1.
and so on, then it's the same way you did.
(a^5)^402 + (a^5)^404 + 1 = 3.
2a^20 1a^10 a^5^5a^2^3^2^3 a^2^1^1^3 a^2^3 (a ➖ 3a+2). 2a^8060 2a^8^6 2a^2^3^2^3 1a^1^1^1^2^3a^2^3(a ➖ 3a+2).
is it 3?
just watch the video
Yes