2023/024) show that $(x+1)^7−x^7−1$ is divisible by $x(x+1)(x^2+x+1)$

First we note that $gcd(x,x+1) = 1$

$gcd(x, x^2 + x + 1) = GCD(x, x(x+1)) = 1$

and $gcd(x+1, x^2 + x + 1) = GCD(x+1, x(x+1)) = 1$

So $x,x+1,x^2+x+1$ are pairwise co-prime and to prove that the given expression is divisible by  $x(x+1)(x^2+x+1)$  we need to show that it is divisible by $x$, $x+1$ and $x^2+x+1$

Now Let $P(x) = (x+1)^7−x^7−1$

To show that it is dvisible by x we need to show P(0) = 0

$P(0) = 1^7 - 0^7 -1 = 0$  so it is divisibe by x

To show that it is dvisible by x+1 we need to show P(-1) = 0

$P(-1) = (-1+1)^7 - (-1)7 -1 = 0$  so it is divisibe by x+1

Now we need to show that it is divisible by $x^2+x+1$

That is if $x^2+x+1=0$ then $P(x) = 0$

Because 

$x^2+x+1=0\cdots(1)$ we get

$x+1 = - x^2\cdots(2)$

And multiplying by $x-1$ we have $x^3-1 = 0$ 

Or $x^3 = 1\cdots(3)$

Now $P(x) = (x+1)^7−x^7−1$

$= (-x^2)^7 - x^7 -1$ using (1)

$=x^{14} - x^7 -1$

$= -(x^3)^4 * x^2 - (x^3)^2 * x - 1$ using (2)

$= -x^2 - x - 1$

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

So P(x) =0 if $x^2+x+1=0$

Hence $x^2+x+1$ is a factor

As $x,x+1,x^2+x+1$ are factors and are pairwise co-prime so   $(x+1)^7−x^7−1$  is divisible by $x(x+1)(x^2+x+1)$