If a, b, are the roots of the equation x^2 + x + 1 calculate a^1980 + b^1980

If x^2+x+1  = 0, then, if we'll multiply both sides by
(x-1),we'll get:

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

But the product (x-1)(x^2 + x + 1) is the difference of
cubes:

(x-1)(x^2 + x + 1) = x^3 -
1

So, if (x-1)(x^2 + x + 1) = 0, then x^3 - 1 =
0

x^3 = 1

Now, if a and b are
the roots of the equation x^2 + x + 1, then they are the roots of the equation x^3 - 1 =
0, too.

According to the rule, each root substituted in the
original equation, verifies it.

a^3 - 1 =
0

a^3 = 1

b^3 - 1 =
0

b^3 = 1

Now, we'll try to
express the exponent 1980 as a multiple of 3:

1980 =
3*660

a^1980 + b^1980 = (a^3)^660 +
(b^3)^660

But a^3 and b^3 =
1

a^1980 + b^1980 = 1^660 +
1^660

a^1980 + b^1980 = 1 +
1

a^1980 + b^1980 =
2