How do you simplify #(x-3)/(x-5) div (x^3-27)/(x^3+8)*(x^2-25)/(x + 5)#?

1 Answer
Mar 15, 2016

Answer:

#((x+2)(x^2-2x+4))/(x^2+3x+9)#

Explanation:

Recall the following:

#1#. Difference of cubes: #color(orange)((a^3-b^3)=(a-b)(a^2+ab+b^2))#

#2#. Sum of cubes: #color(blue)((a^3+b^3)=(a+b)(a^2-ab+b^2))#

#3#. Difference of squares: #color(purple)((a^2-b^2)=(a+b)(a-b))#

The Simplification Process
#1#. Start by factoring the polynomials in the numerator and denominator of the second fraction.

#(x-3)/(x-5)-:(x^3-27)/(x^3+8)*(x^2-25)/(x+5)#

#=(x-3)/(x-5)-:(color(orange)((x-3)((x)^2+(3)(x)+(3)^2)))/(color(blue)((x+2)((x)^2-(2)(x)+(2)^2)))*(x^2-25)/(x+5)#

#2#. Simplify.

#=(x-3)/(x-5)-:(color(orange)((x-3)(x^2+3x+9)))/(color(blue)((x+2)(x^2-2x+4)))*(x^2-25)/(x+5)#

#3#. Factor the polynomials in the numerator and denominator of the third fraction.

#=(x-3)/(x-5)-:(color(orange)((x-3)(x^2+3x+9)))/(color(blue)((x+2)(x^2-2x+4)))*(color(purple)((x+5)(x-5)))/(x+5)#

#4#. To divide the first fraction by the second fraction, rewrite the second fraction as its reciprocal.

#=(x-3)/(x-5)*(color(blue)((x+2)(x^2-2x+4)))/(color(orange)((x-3)(x^2+3x+9)))*color(purple)((x+5)(x-5))/(x+5)#

#5#. Cancel out any factors which appear in the numerator and denominator as a pair.

#=color(red)cancelcolor(black)((x-3))/color(teal)cancelcolor(black)((x-5))*(color(blue)((x+2)(x^2-2x+4)))/(color(orange)(color(red)cancelcolor(orange)((x-3))(x^2+3x+9)))*color(purple)(color(magenta)cancelcolor(purple)((x+5))color(teal)cancelcolor(purple)((x-5)))/color(magenta)cancelcolor(black)((x+5))#

#6#. Rewrite the expression.

#=color(green)(|bar(ul(color(white)(a/a)((x+2)(x^2-2x+4))/(x^2+3x+9)color(white)(a/a)|)))#