Question 43475
 
 {{{(8y^3+27)/(2xy-10y+3x-15)}}}={{{((2y)^3+3^3)/(2xy+3x-10y-15)}}}
 = {{{(2y+3)(4y^2-6y+9)/(x(2y+3)-5(2y+3))}}}={{{(2y+3)(4y^2-6y+9)/(2y+3)(x-5)}}}={{{cross((2y+3))(4y^2-6y+9)/cross((2y+3))(x-5)}}}
=  {{{(4y^2-6y+9)/(x-5)}}} Answer.

Is it clear now ?

gsm