If $\frac{x^{3}-1}{p(x)}=\frac{x^{2}+x+1}{x-1}$,then $p(x) = \dots$

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following:
$p(x) = x^{4} - 5x + 6, \quad g(x) = 2 - x^{2}$

$(x+1)$ is not one of the factors of .........

What must be added to the polynomial $p(x) = 6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$ so that the resulting polynomial is exactly divisible by $3x^2 - 2x + 4$? (Hint: add $-r(x)$ to $p(x)$)

Divide $x^{3}-6 x^{2}+11 x-6$ by $x^{2}+x+1$.

If the polynomial $p(x) = x^{3} - 3x^{2} + x + 2$ is divided by the divisor polynomial $s(x)$, then the quotient polynomial $q(x) = x - 2$ and the remainder polynomial $r(x) = -2x + 4$ are obtained. Find the divisor polynomial $s(x)$.