SOLUTION: factor by grouping {{{x^3+y^3-x^2y-xy^2}}}

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Question 1089262: factor by grouping
Found 4 solutions by josgarithmetic, Theo, ikleyn, MathTherapy:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website!

not finished...

Answer by Theo(13342)   (Show Source):
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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.

Answer by ikleyn(52805)   (Show Source):
You can put this solution on YOUR website!
.
The solution and the answer by @josgarithmetic is WRONG.

Answer by MathTherapy(10553)   (Show Source):
You can put this solution on YOUR website!

factor by grouping

Correct answer:
IGNORE all other WRONG answers.