If x1 and x2 are the roots of equation x^2+x+1=0, find x1^10+x2^10=?

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

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

But the product is the result of difference of
cubes:

x^3 - 1 = (x-1)(x^2 + x +
1)

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

We'll add 1 both sides:

x^3
= 1

x1 and x2 are the roots of the equation x^3 - 1 = 0,
so:

x1^3 = 1

x2^3  =
1

We'll write x1^10 =
x1^(9+1)

x1^(9+1)
= x1^9*x1

x1^9*x1 =
(x1^3)^2*x1

But x1^3 = 1

x1^10
= 1^2*x1

x1^10 = x1

x2^10 =
x2

So, x1^10  + x2^10 = x1 +
x2

We'll use Viete's relations to express x1 +
x2:

x1 + x2 = -b/a

where b and
a are the coefficients of the quadratic equation

ax^2 + bx
+ c = 0

In our case, the quadratic
is:

x^2 +x +1 = 0

a =
1

b = 1

x1 + x2 =
-1/1

x1 + x2 =
-1

x1^10  + x2^10 =
-1