SOLUTION: how do you factor x to the 6 minus y to the 6th two different ways. I think one way would be to use the difference of squares and the other the difference of cubes.

Algebra ->  Exponents -> SOLUTION: how do you factor x to the 6 minus y to the 6th two different ways. I think one way would be to use the difference of squares and the other the difference of cubes.       Log On


   

Question 664994: how do you factor x to the 6 minus y to the 6th two different ways. I think one way would be to use the difference of squares and the other the difference of cubes.
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
correct
Let's do diff of squares first. For that
a%5E2+-+b%5E2%29+=+%28a+%2B+b%29%28a-b%29
In our case, a^2 is x^6, so a = x^3. b^2 is y^6, so b - y^3
thus we get
%28x%5E3%29%5E2+-+%28y%5E3%29%5E2+=+%28x%5E3+%2B+y%5E3%29%28x%5E3+-+y%5E3%29 note terms are also a sum and diff of 2 cubes, so you need to expand those as well. End result is

Now let's do it as diff of two cubes first
%28x%5E2%29%5E3+-+%28y%5E2%29%5E3+=+%28x%5E2+-+y%5E2%29%28x%5E4+%2Bx%5E2y%5E2+%2B+y%5E4%29 but the first term is a diff of 2 squares, so expand that to
%28x-y%29%28x%2By%29%28x%5E4+%2Bx%5E2y%5E2+%2B+y%5E4%29 i do not have a formula that shows how to easily expand the last term, but since hte factoring must come out the same regardless lf the method used, you can prove to yourself that
%28x-y%29%28x%2By%29%28x%5E2+-xy+%2B+y%5E2%29%28x%5E2+%2B+xy+%2B+y%5E2%29