Factor x^15+y^15 consider it sum of two fifths powers?

Question

Factor x^15+y^15 by first considering it to be a sum of two fifths powers.
Please explain in steps how to do this. I'll give best answer.

Thank you so much Doc B! I'll give you best answer as soon as I can. I wish all answers are as good as yours.

Answer 1

x^15 + y^15 =

(x^3)^5 + (y^3)^5

now you need to know the formula for factoring a^5 + b^5
which I, unfortunately, don't know

Answer 2

Suppose you have a polynomial of the form x^n + y^n, where n is odd. If it should happen that y=-x, then the whole thing reduces to zero. That is, when x+y=0, the whole polynomial is zero. Therefore, (x+y) is a factor of the polynomial.

Example: x^3 + y^3 = (x+y)(x^2 - xy + y^2)
That's the "Sum of two cubes" formula.

Here's the "Sum of two fifth powers":
x^5 + y^5 = (x+y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4).

The first factor is just the sum of the base variables. The other factor alternates +, -, +, -, +, while exchanging an x for a y with each term. (x^4 becomes x^3y, etc.)

When we factor the sum of two 15th powers, we could extend the same pattern, or we could exploit the fact that 15=3*5.

I think the easiest way to do that is to let u=x^3, v=y^3.
Now u^5=x^15 and v^5=y^15, so
x^15 + y^15 = u^5 + v^5
= (u+v)(u^4 - u^3v + u^2v^2 - uv^3 + v^4)
= (x^3 + y^3)(x^12 - x^9y^3 + ... + y^12)

You may want to factor a little further, using the sum of two cubes on that leading factor...
=(x+y)(x^2 - xy + y^2)(x^12 - ... + y^12)

This can be factored further, but I think our job is done.

Answer 3

the first step: Divide 20 via 5, this provides one fifth. Step 2: you opt for 2 fifths, so take one fifth and double it. So truly 20 divided via 5 is 4, so 4 is one fifth. 4 doubled is 8. So 8 is two fifths of 20.