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

(N/A) To divide $x^{3}-6 x^{2}+11 x-6$ by $x^{2}+x+1$,we use the long division method:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x^{2})$ to get $x$.
$2$. Multiply $x$ by the divisor $(x^{2}+x+1)$ to get $x^{3}+x^{2}+x$.
$3$. Subtract this from the dividend: $(x^{3}-6 x^{2}+11 x-6) - (x^{3}+x^{2}+x) = -7 x^{2}+10 x-6$.
$4$. Divide the first term of the new polynomial $(-7 x^{2})$ by the first term of the divisor $(x^{2})$ to get $-7$.
$5$. Multiply $-7$ by the divisor $(x^{2}+x+1)$ to get $-7 x^{2}-7 x-7$.
$6$. Subtract this from the current polynomial: $(-7 x^{2}+10 x-6) - (-7 x^{2}-7 x-7) = 17 x+1$.
Thus,the quotient is $x-7$ and the remainder is $17 x+1$.