Calculate (z1+1)^10+(z2+1)^10. z1, z2 are the solutions of equation z^2+z+1.

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

(z-1)(z^2 + z + 1) =
0

But the product is the result of difference of
cubes:

z^3 - 1 = (z-1)(z^2 + z +
1)

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

We'll add 1 both sides:

z^3
= 1

z1 and z2 are the roots of the equation z^3 - 1 = 0,
so:

z1^3 = 1

z2^3  =
1

From the equation z^2 + z + 1 =
0

z + 1 = -z^2

We'll
substitute z1 and z2:

z1 + 1 =
-z1^2

z2 + 1 = -z2^2

(z1 +
1)^10 + (z2 + 1)^10 = (-z1^2)^10 + (-z2^2)^10

We'll
re-write the expression (-z1^2)^10 + (-z2^2)^10:

(-z1^2)^10
+ (-z2^2)^10 = (z1^2)^10 + (z2^2)^10

(z1^10)^2 +
(z2^10)^2

We'll write z1^10 =
z1^(9+1)

z1^(9+1)
= z1^9*z1

z1^9*z1 =
(z1^3)^2*z1

But z1^3 = 1

z1^10
= 1^2*z1

z1^10 = z1

z2^10 =
z2

So, 

(z1^10)^2 + (z2^10)^2
= z1^2 + z2^2

We'll use Viete's relations to express z1 +
z2:

z1 + z2 = -b/a

where b and
a are the coefficients of the quadratic equation

az^2 + bz
+ c = 0

In our case, the quadratic
is:

z^2 +z +1 = 0

a =
1

b = 1

c =
1

z1 + z2 = -1/1

z1 + z2 =
-1

z1*z2 = c/a

z1*z2 =
1

z1^2 + z2^2 = (z1 + z2)^2 -
2z1*z2

(z1 + 1)^10 + (z2 + 1)^10 = (-1)^2 -
2*1

(z1 + 1)^10 + (z2 + 1)^10 = 1 -
2

(z1 + 1)^10 + (z2 + 1)^10 =
-1