How do you divide #(x^3-1)/(x^2+1) div (9x^2+9x+9)/(x^2-x)#? Algebra Rational Equations and Functions Division of Rational Expressions 1 Answer Sonnhard Jun 22, 2018 #x(x-1)/9# Explanation: Note that #x^3-1=(x-1)(x^2+x+1)# we get #(x^3-1)/(x-1)*(x(x-1))/(9(x^2+x+1))#= #((x-1)(x^2+x+1)*x*(x-1))/(9(x^2+x+1)(x-1))=(x(x-1))/9# Answer link Related questions What is Division of Rational Expressions? How does the division of rational expressions differ from the multiplication of rational expressions? How do you divide 3 rational expressions? How do you divide rational expressions? How do you divide and simplify #\frac{9x^2-4}{2x-2} -: \frac{21x^2-2x-8}{1} #? How do you divide and reduce the expression to the lowest terms #2xy \-: \frac{2x^2}{y}#? How do you divide #\frac{x^2-25}{x+3} \-: (x-5)#? How do you divide #\frac{a^2+2ab+b^2}{ab^2-a^2b} \-: (a+b)#? How do you simplify #(w^2+6w+5)/(w+5)#? How do you simplify #(x^4-256)/(x-4)#? See all questions in Division of Rational Expressions Impact of this question 1490 views around the world You can reuse this answer Creative Commons License