Question 56058
:
assume you mean:
{{{(x^2+2x)/(4x-3)}}}
---------------
{{{(6x^3)/(8x^2-2x-3)}}}
:
Invert the dividing fraction and mult;
{{{(x^2+2x)/(4x-3)}}} * {{{(8x^2 - 2x -  3)/(6x^3)}}}
:
{{{(x^2+2x)/(4x-3)}}} * {{{(4x-3)(2x-1)/(6x^3)}}}
:
Cancel the (4x-3)'s
{{{(x^2+2x) * (2x-1)/(6x^3)}}} = {{{(2x^3 + 3x^2 - 2x)/(6x^3)}}} = {{{(x(2x^2 + 3x - 2))/(6x^3)}}} = {{{(2x^2 + 3x - 2)/(6x^2)}}} = {{{((2x-1)(x+2))/(6x^2)}}}