How do you simplify #(r^2+25) / (8r^3 -27) - ( 3r+2) /(2r-3)#?

Question

1348 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-25T14:42:52

#(-12r^3-25r^2-39r+7)/[(2r-3)(4r^2+6r+9)#

First, factor #8r^3-27# into #(2r-3)(4r^2+6r+9)# using #(a^3-b^3)=(a-b)(a^2+ab+b^2)#.

We now get

#(r^2+25)/[(2r-3)(4r^2+6r+9)]-(3r+2)/(2r-3)#

Multiply by #(4r^2+6r+9)# in the second expression to get

#[(3r+2)(4r^2+6r+9)]/[(2r-3)(4r^2+6r+9)#

Now you can combine to get

#{(r^2+25)-(3r+2)(4r^2+6r+9)}/[(2r-3)(4r^2+6r+9)#

Expanding gives you

#{(r^2+25)-(12r^3+26r^2+39r+18)}/[(2r-3)(4r^2+6r+9)#

Last step is to simplify and you will get

#(-12r^3-25r^2-39r+7)/[(2r-3)(4r^2+6r+9)#