Question 165732
{{{(8y^3+27)/(2xy-10y+3x-15)}}} Start with the given expression



{{{((2y+3)(4y^2-6y+9))/(2xy-10y+3x-15)}}} Factor the numerator. Note: use the <a href="http://www.purplemath.com/modules/specfact2.htm">Sum of Cubes formula</a> to factor.



{{{((2y+3)(4y^2-6y+9))/((2y+3)(x-5))}}} Factor the denominator.



{{{(highlight((2y+3))(4y^2-6y+9))/(highlight((2y+3))(x-5))}}} Highlight the common terms.



{{{(cross((2y+3))(4y^2-6y+9))/(cross((2y+3))(x-5))}}} Cancel out the common terms.



{{{(4y^2-6y+9)/(x-5)}}} Simplify



So {{{(8y^3+27)/(2xy-10y+3x-15)}}} simplifies to {{{(4y^2-6y+9)/(x-5)}}}