Divide $x^{3}-6x^{2}+11x-6$ by $x^{2}-8x+27$.

(N/A) To divide $x^{3}-6x^{2}+11x-6$ by $x^{2}-8x+27$,we use polynomial long division:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x^{2})$: $x^{3} / x^{2} = x$.
$2$. Multiply the divisor $(x^{2}-8x+27)$ by $x$: $x(x^{2}-8x+27) = x^{3}-8x^{2}+27x$.
$3$. Subtract this from the dividend: $(x^{3}-6x^{2}+11x-6) - (x^{3}-8x^{2}+27x) = 2x^{2}-16x-6$.
$4$. Divide the first term of the new polynomial $(2x^{2})$ by the first term of the divisor $(x^{2})$: $2x^{2} / x^{2} = 2$.
$5$. Multiply the divisor $(x^{2}-8x+27)$ by $2$: $2(x^{2}-8x+27) = 2x^{2}-16x+54$.
$6$. Subtract this from the current polynomial: $(2x^{2}-16x-6) - (2x^{2}-16x+54) = -60$.
Thus,the quotient is $x+2$ and the remainder is $-60$.