And the 3rd one:


\({3x(2x-3)-(3-2x) \over 3-2x}-{18x^3 \over 9x^3-27x^2+x-3}\)


This is a lenghty and complicated sum, to try and add here the steps I took, will take a looong time, if someone out there could show me the way with this one also, I would really appreciate it!!

juriemagic  Nov 1, 2017

I do not think this simplifies very nicely


\({3x(2x-3)-(3-2x) \over 3-2x} - {18x^3 \over 9x^3-27x^2+x-3}\\ ={-3x(3-2x)-1(3-2x) \over 3-2x} - {18x^3 \over 9x^2(x-3)+1(x-3)}\\ ={(-3x-1)(3-2x) \over 3-2x} - {18x^3 \over (9x^2+1)(x-3)}\\ ={(-3x-1)(3-2x) \over 3-2x} - {18x^3 \over (9x^2+1)(x-3)}\\ ={-(3x+1)} - {18x^3 \over (9x^2+1)(x-3)}\\ ={-(3x+1) (9x^2+1)(x-3)-18x^3 \over (9x^2+1)(x-3)}\\\)


You can expand the numerator and denominator out if you want but nothing will cancel.


I should have added that x cannot equal 3 or 3/2

Melody  Nov 1, 2017
edited by Melody  Nov 1, 2017

The others were equations .. this one is a subtraction.

Maybe it was meant to be an equals sign ?

Melody  Nov 1, 2017

Hi Melody,


the left side I also got, however this is what I did with the right side:

\({18x^3} \over9x^3+x-27x^2-3\)

\(18x^3 \over(9x^3+x)+(-27x^2-3)\)

\(18x^3 \over x(9x^2+1)-3(9x^2+1)\)

\(18x^3 \over (x-3)(9x^2+1)\)

\(18x^2 \over (x-3)(3x+1)(3x+1)\)


so now it's;


\(-(3x+1)-{18x^3 \over(x-3)(3x+1)(3x+1)}\)


and this is where it ends for me...yes, I do not know, either = or * would have worked nicely, but subtraction?..I don't know..


Melody, anyways, thank you kindly once again for your help...I think I'll let this one pass..

juriemagic  Nov 1, 2017

Hi Juriemagic,

Your factorisation on the right is not correct.


the first few lines are good but


\(9x^2+1 \ne (3x+1)^2\\~\\ (3x+1)^2=9x^2+6x+1\)



Other than that little error our answers are the same I think :)

Melody  Nov 1, 2017


Yep, I see now...Thank you!!

juriemagic  Nov 2, 2017

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