Question 1089262
i think this would be factored as follows.


start with x^3 + y^3 - x^2y - xy^2


rearrange by descending order of degree of x.


you get:


x^3 - x^2y - xy^2 + y^3


group as shown below:


(x^3 - x^2y) - (xy^2 - y^3)


on the left group, factor out x^2.


on the right group, factor out y^2.


you will get:


x^2 * (x - y) - y^2 * (x - y)


factor out the common term of (x - y) and you get:


(x^2 - y^2) * (x - y)


(x^2 - y^2) is equal to (x - y) * (x + y)


your final factored expression becomes:


(x - y) * (x + y) * (x - y)


this expression should be equivalent to x^3 + y^3 - x^2y - xy^2


to see if it is, pick a random value for x and y and see if both expressions give you the same result.


i'll pick x = 2 and y = 3.


x^3 + y^3 - x^2y - xy^2 is equal to 5


(x-y) * (x+y) * (x-y) is equal to 5 as well.


it looks like these expressions are equivalent.


this is reasonable proof that the factorization was correct.


you can try it with other random values of x and y and you should see that the expressions are equivalent.


i redid the experiment with x = 7 and y = 4


i got 99 using each expression.


that solidifies the impression that the factorization was done correctly.