Prove that the polynomial (x^2+x+1)^(8n+1) -x is divisible by x^2+1.

To prove that the polynomial (x^2+x+1)^(8n+1) -x is
divisible by x^2 + 1, we'll have to write the given polynomial as a product of factors
and one of these factors to be a multiple or even the divisor x^2 +
1.

We'll put x^2 + 1 = 0

We'll substitute x^2 + 1 by
the value 0 into the brackets:

(x^2+x+1)^(8n+1) - x = x^(8n+1) -
x

But x^(8n+1) = x^(8n)*x

x^(8n+1) - x = x^(8n)*x -
x

We'll factorize by x:

x^(8n)*x - x = x*[x^(8n) -
1]

We notice that inside the brackets we have a difference of 2
squares:

x*[x^(8n) - 1] = x*[x^(4n) - 1]*[x^(4n) +
1]

x*[x^(4n) - 1]*[x^(4n) + 1] = x*[x^(2n) - 1]*[x^(2n) + 1]*[x^(4n) +
1]

As we can notice, one of these factors is x^(2n) + 1, that is
cancelling.

x*[x^(2n) - 1]*[x^(2n) + 1]*[x^(4n) + 1] = x*[x^(2n) -
1]*0*[x^(4n) + 1] = 0

Since, replacing into the given
polynomial the divisor x^2+1, it yields 0, then the polynomial (x^2+x+1)^(8n+1) -x is
divisible by x^2 + 1.