# if a and b are roots of x^2+x+1, what is a^1980+b^1980?

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### 2 Answers

Multiplying the equation `x^2+ x + 1 = 0` , both sides by `(x-1), ` yields:

`(x - 1)(x^2+ x + 1) = (x - 1)*0`

Notice that the product from left side of equal my be replaced by difference of cubes such that:

`x^3 - 1 = 0 =gt x^3 = 1 =gt x = 1`

`` Hence, if a root of the equation `x^3 - 1 = 0` is `x = 1` , then you may as well plug this root in the addition of powers.

`a^1980 + b^1980 = 1^1980 + 1^1980 = 1 + 1 = 2`

**Hence, the sum of powers `a^1980 + b^1980` yields 2.**

a and b are roots of x^2 + x + 1

The roots of x^2 + x + 1 are:

a = `(-1+sqrt(1 - 4))/2 = -1/2 + i*sqrt 3/2`

b = `(-1 - sqrt(1 - 4))/2 = -1/2 - i*sqrt 3/2`

a^2 = ` `-a - 1

a^4 = -a - 1 + 1 + 2a = a

a^1980 = a^495 = a^3*a^123 = a^3*a^3*a^30 = a^6*a^2*a^7 = a^2*a^7 = a^3

Similarly b^1980 = b^3

a^1980 + b^1980 = a^3 + b^3

=> 2

**The required value of a^1980 + b^1980 = 2**