# Factoring the Sum of Two Fifth Powers

1. Jun 18, 2008

My problem is to factor $$x^{5}$$ + $$y^{5}$$.

I get $$(x+y)(x^{8}-x^{4}y^{4}+y^{8})$$.

However, the answer is $$(x+y)(x^{4}-x^{2}y^{2}+y^{4})$$.

2. Jun 18, 2008

### mathman

Neither is right! Multiplying out in either cases does not give you the original back.

3. Jun 19, 2008

$$x^{5}+y^{5}$$
$$=x^{3}x^{2}+y^{3}y^{2}$$​

$$=(x^{2}x+y^{2}y)(x^{4}x^{2}-x^{2}xy^{2}y+y^{4}y^{2})$$​

$$=(x+y)(x^{2}-xy+y^{2})(x^{6}-x^{3}y^{3}+y^{6})$$​

$$=(x+y)(x^{8}-x^{5}y^{3}+x^{2}y^{6}-x^{7}y+x^{4}y^{3}-xy^{7}+x^{6}y^{2}-x^{3}y^{5}+y^{8})$$​

What am I doing wrong?

------------------------

$$=(x+y)(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4})$$​

4. Jun 19, 2008

### Hurkyl

Staff Emeritus
The first and last steps are the only ones that make any sense; I have no idea what you were trying to do inbetween.

5. Jun 19, 2008

### Werg22

How about you try factoring 5*x^4*y + 10*x^3*y^2 + 10* x^2 * y^3 + 5*x*y^4 instead? This gives 5*xy( (x+y)^3 - x^2*y - x*y^2) = 5*xy( (x+y)^3 - xy(x + y)) = 5*xy*(x+y)*((x+y)^2 - xy). Since x^5 + y^5 = (x+y)^5 - (5*x^4*y + 10*x^3*y^2 + 10* x^2 * y^3 + 5*x*y^4), we obtain x^5 + y^5 = (x+y)^5 - 5*xy*(x+y)*((x+y)^2 - xy) = (x+y)((x+y)^4 - 5*xy*(x+y)*((x+y)^2 - xy)), and take it from there (you can extract (x+y) still).

Last edited: Jun 19, 2008
6. Jun 19, 2008

### Gib Z

You could think of that expression as [/itex]x^5 - (-y)^5[/itex] and apply that into $$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-1})$$. You can verify that general equation by expanding the RHS.

7. Jun 19, 2008

### mathman

Your last assertion makes no sense. You have terms x9 and y9 among others. They seem to come out of nowhere!!!!

8. Jun 19, 2008

I thought applying the sum of cubes on the cubed power parts would have worked $$x^{3}x^{2}+y^{3}y^{2}$$.