What is t if x(x+1)^(2n+1)+(t-1)x^n is divided by x^2+x+1?

If x(x+1)^(2n+1)+(t-1)x^n is divided by x^2+x+1, then the
roots of  x^2+x+1 cancel the expression
x(x+1)^(2n+1)+(t-1)x^n.

The roots of x^2+x+1 are found when
x^2+x+1 = 0.

We'll multiply x^2+x+1 by x -
1:

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

We'll
get the difference of cubes:

x^3 - 1 = 0 => x^3 =
1

We'll write the factor x + 1 = -x^2 (from x^2+x+1 =
0)

Replacing x + 1 by - x^2, the polynomial will
become:

x*(-x^2)^2n*(-x^2) + (t-1)*x^n =
0

-x*x^2 = -x^3 = -1

-1*x^4n +
(t-1)*x^n = 0

But x^4n =
(x^3n)*(x^n)

-(x^3n)*(x^n) + (t-1)*x^n =
0

-[(x^3)^n]*(x^n) + (t-1)*x^n =
0

We'll factorize by
x^n:

x^n*(-1 + t - 1) = 0

t -
2 = 0

t = 2

The
value of t, when x(x+1)^(2n+1)+(t-1)x^n is divided by x^2+x+1, is t =
2.