Question 196861
Note: the previous solution has a sign error and the answer should be much simpler.




{{{((x^3+y^3)/(2x^3+2x^2y))/((3x^3-3x^2y+3xy^2)/(6x^2-6y^2))}}} Start with the given expression.



{{{((x^3+y^3)/(2x^3+2x^2y))((6x^2-6y^2)/(3x^3-3x^2y+3xy^2))}}} Multiply the first fraction {{{(x^3+y^3)/(2x^3+2x^2y)}}} by the reciprocal of the second fraction {{{(3x^3-3x^2y+3xy^2)/(6x^2-6y^2)}}}.



{{{(((x+y)(x^2-xy+y^2))/(2x^3+2x^2y))((6x^2-6y^2)/(3x^3-3x^2y+3xy^2))}}} Factor {{{x^3+y^3}}} to get {{{(x+y)(x^2-xy+y^2)}}}.



{{{(((x+y)(x^2-xy+y^2))/(2x^2(x+y)))((6x^2-6y^2)/(3x^3-3x^2y+3xy^2))}}} Factor {{{2x^3+2x^2y}}} to get {{{2x^2(x+y)}}}.



{{{(((x+y)(x^2-xy+y^2))/(2x^2(x+y)))((6(x-y)(x+y))/(3x^3-3x^2y+3xy^2))}}} Factor {{{6x^2-6y^2}}} to get {{{6(x-y)(x+y)}}}.



{{{(((x+y)(x^2-xy+y^2))/(2x^2(x+y)))((6(x-y)(x+y))/(3x(x^2-xy+y^2)))}}} Factor {{{3x^3-3x^2y+3xy^2}}} to get {{{3x(x^2-xy+y^2)}}}.



{{{(6(x+y)(x^2-xy+y^2)(x-y)(x+y))/(2x^2*3x(x+y)(x^2-xy+y^2))}}} Combine the fractions.



{{{(6(x+y)(x^2-xy+y^2)(x-y)(x+y))/(6x^3(x+y)(x^2-xy+y^2))}}} Multiply {{{2x^2}}} and {{{3x}}} to get {{{6x^3}}}.



{{{(highlight(6)*highlight((x+y))highlight((x^2-xy+y^2))(x-y)(x+y))/(highlight(6)x^3*highlight((x+y))highlight((x^2-xy+y^2)))}}} Highlight the common terms. 



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



{{{((x-y)(x+y))/(x^3)}}} Simplify



{{{(x^2-y^2)/(x^3)}}} FOIL




So {{{((x^3+y^3)/(2x^3+2x^2y))/((3x^3-3x^2y+3xy^2)/(6x^2-6y^2))}}} simplifies to {{{(x^2-y^2)/(x^3)}}}.



In other words, {{{((x^3+y^3)/(2x^3+2x^2y))/((3x^3-3x^2y+3xy^2)/(6x^2-6y^2))=(x^2-y^2)/(x^3)}}}