Factoring x^6 - y^6 as a difference of squares vs cubes?

I agree that understanding the difference of two squares is important, but the problem statement is asking to factor x6 - y6, which is the difference of two cubes.In summary, the conversation discusses the factorization of x6 - y6 and how to factor it into (x2 + xy + y2)(x2 - xy + y2). It is revealed that the polynomial x4 + x2y2 + y4 can also be factored using the difference of two squares. Various online factorers are mentioned, with the recommendation of using WolframAlpha for more advanced problems.
  • #1
Esoremada
52
0

Homework Statement



Factor x6 - y6

Homework Equations



a3 - b3 = (a - b)(a2 + ab + b2)
a2 - b2 = (a + b)(a - b)

The Attempt at a Solution



I'm confused.

x6 - y6 = (x2)3 - (y2)3 = (x3)2 - (y3)2

So shouldn't they all have the same factors? When I factor (x2)3 - (y2)3 = (x3)2 - (y3)2 I get different results.

(x3)2 - (y3)2
(x3 - y3)(x3 + y3)
(x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)(x2)3 - (y2)3
(x2 - y2)((x2)2 + x2y2 + (y2)2)
(x – y)(x + y)(x4 + x2y2 + y4)How do you factor (x4 + x2y2 + y4) into (x2 + xy + y2)(x2 - xy + y2)? Online factorers are saying it's not possible.
 
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  • #2
Esoremada said:
How do you factor (x4 + x2y2 + y4) into (x2 + xy + y2)(x2 - xy + y2)? Online factorers are saying it's not possible.

x4 + x2y2 + y4= (x4 + 2x2y2 + y4)-x2y2

ehild
 
  • #3
Esoremada said:

Homework Statement



Factor x6 - y6

Homework Equations



a3 - b3 = (a - b)(a2 + ab + b2)
a2 - b2 = (a + b)(a - b)

The Attempt at a Solution



I'm confused.

x6 - y6 = (x2)3 - (y2)3 = (x3)2 - (y3)2

So shouldn't they all have the same factors? When I factor (x2)3 - (y2)3 = (x3)2 - (y3)2 I get different results.

(x3)2 - (y3)2
(x3 - y3)(x3 + y3)
(x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)(x2)3 - (y2)3
(x2 - y2)((x2)2 + x2y2 + (y2)2)
(x – y)(x + y)(x4 + x2y2 + y4)How do you factor (x4 + x2y2 + y4) into (x2 + xy + y2)(x2 - xy + y2)? Online factorers are saying it's not possible.

A little something I also discovered for myself back in high school. I assumed it was a closely guarded secret that only exceptional Mathematicians discover in their lifetimes :wink:
 
  • #4
You are an exceptional Mathematician then... But our teacher told us in the school that adding and subtracting the same thing does not hurt.:smile:

ehild
 
  • #5
Well, if you work out the product (x^2 + xy + y^2)(x^2 -xy + y^2) you get the right result. What else do you want?
 
  • #6
M Quack said:
Well, if you work out the product (x^2 + xy + y^2)(x^2 -xy + y^2) you get the right result. What else do you want?

To find out how to do the factoring if you do not know that these are the factors.

As I wrote before, x4 + x2y2 + y4= (x4 + 2x2y2 + y4)-x2y2=(x2+y2)2-(xy)2.
Apply the identity a2-b2=(a-b)(a+b)


ehild
 
  • #7
Esoremada said:

Homework Statement



Factor x6 - y6


Homework Equations



a3 - b3 = (a - b)(a2 + ab + b2)
a2 - b2 = (a + b)(a - b)

The Attempt at a Solution



I'm confused.

x6 - y6 = (x2)3 - (y2)3 = (x3)2 - (y3)2

So shouldn't they all have the same factors? When I factor (x2)3 - (y2)3 = (x3)2 - (y3)2 I get different results.

(x3)2 - (y3)2
(x3 - y3)(x3 + y3)
(x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)


(x2)3 - (y2)3
(x2 - y2)((x2)2 + x2y2 + (y2)2)
(x – y)(x + y)(x4 + x2y2 + y4)


How do you factor (x4 + x2y2 + y4) into (x2 + xy + y2)(x2 - xy + y2)? Online factorers are saying it's not possible.

What online factorers are you using? Certainly Maple can do it with no problem (although it is not an on-line package_.

RGV
 
  • #9
Esoremada said:
http://www.freemathhelp.com/factoring-calculator.php

(x^4 + x^2y^2 + y^4)

The polynomial is not factorable with real numbers.
That's what the website says

To paraphrase Groucho Marx: What you gonna' believe, that website or your lyin' eyes?
 
Last edited:
  • #10
I understand now, thanks for the help :smile:

Not going to trust that website as much anymore
 
  • #11
Esoremada said:
How do you factor (x4 + x2y2 + y4)? Online factorers are saying it's not possible.

ehild said:
x4 + x2y2 + y4= (x4 + 2x2y2 + y4) - x2y2

What ehild has shown is the difference of two squares, not factors. Note the problem statement is asking for the difference of two squares.
 
Last edited:
  • #12
Esoremada said:
I understand now, thanks for the help :smile:

Not going to trust that website as much anymore

You should try http://www.wolframalpha.com/
It can handle much more advanced problems.
 
  • #14
rcgldr said:
What ehild has shown is the difference of two squares, not factors. Note the problem statement is asking for the difference of two squares.

@rcgldr
I must not give full solution. It was a hint. The difference of two squares is easy to factorize.

hild
 
  • #15
ehild said:
@rcgldr - I must not give full solution. It was a hint. The difference of two squares is easy to factorize.
I somehow missed post #6 where you explained this.
 
Last edited:

1. What is the difference between factoring x^6 - y^6 as a difference of squares vs cubes?

The main difference between factoring x^6 - y^6 as a difference of squares and as a difference of cubes is the number of terms in the final factored expression. When factoring as a difference of squares, the final expression will have two terms, while factoring as a difference of cubes will result in three terms.

2. How do you factor x^6 - y^6 as a difference of squares?

To factor x^6 - y^6 as a difference of squares, you need to use the formula (a^2 - b^2) = (a + b)(a - b). In this case, a = x^3 and b = y^3, so the factored expression would be (x^3 + y^3)(x^3 - y^3).

3. How do you factor x^6 - y^6 as a difference of cubes?

To factor x^6 - y^6 as a difference of cubes, you need to use the formula (a^3 - b^3) = (a - b)(a^2 + ab + b^2). In this case, a = x^2 and b = y^2, so the factored expression would be (x^2 - y^2)(x^4 + x^2y^2 + y^4).

4. Which method is more efficient for factoring x^6 - y^6?

The method of factoring x^6 - y^6 as a difference of squares is more efficient because it only requires one step using the formula, while factoring as a difference of cubes involves an additional step of using the formula for the sum of cubes.

5. Can you use either method for factoring any polynomial of the form x^m - y^n?

No, the methods of factoring as a difference of squares or cubes only work for polynomials of the form x^m - y^n when the exponents m and n are even and have a greatest common divisor of 2. If the exponents are odd or do not have a GCD of 2, then the polynomial cannot be factored using these methods.

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