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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.
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