Using the remainder theorem we can prove that (x^2+x+1)^(8n+1) -x is divisible by x^2+1.

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

f(x) = (x^2+x+1)^(8n+1) - x is divisible by x^2 + 1 if f(i) = 0 and f(-i) = 0

f(i) = (i^2 + i + 1)^(8n+1) - i

=> (-1 + i + 1)^(8n + 1) - i

=> i^8n*i - i

=> i*(i^8n - 1)

=> i*((-1)^4n - 1)

=> i*(1^2n - 1)

=> i*0

=> 0

f( -i)

=> ((-i)^2 - i + 1)^(8n+1) + i

=> (-1 - i + 1)^(8n + 1) + i

=> (-i)^(8n)*-i + i

=> -i*((-i)^8n - 1)

=> -i*((-1)^4n - 1)

=> -i*(1^2n - 1)

=> -i*0

=> 0

This has proved that (x^2+x+1)^(8n+1) - x is divisible by (x - i) and by (x + i)

**Therefore (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.**