MHB X^6 - y^6 As Difference of Squares

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The discussion focuses on factoring the expression x^6 - y^6 as a difference of squares. The initial factorization is presented as (x^3 - y^3)(x^3 + y^3), leading to the application of the difference of cubes for further simplification. The first factor can be factored using the difference of squares, while the second factor is addressed through a system of equations to find coefficients. Ultimately, the expression is fully factored as (x^2 - y^2)(x^4 + x^2y^2 + y^4) and further simplified to (x^2 + xy + y^2)(x^2 - xy + y^2). The discussion emphasizes the importance of recognizing both difference of squares and cubes in the factorization process.
mathdad
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Factor x^6 - y^6 as a difference of squares.

Solution:

(x^3 - y^3)(x^3 + y^3)

The problems states to use the difference of squares.

I can apply the difference of cubes to the left factor and the sum of cubes to the right factor but how do I continue using the difference of squares?
 
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Once you've done that' you've applied the difference of squares...to continue factoring further, as you noted, you need to use the sum/difference of cubes. Now, if you begin with the difference of cubes, you'd have:

$$x^6-y^6=(x^2-y^2)(x^4+x^2y^2+y^4)$$

Now to continue, you would use the difference of squares on the first factor. To factor the second factor, we can try:

$$x^4+x^2y^2+y^4=(x^2+axy+y^2)(x^2+bxy+y^2)=x^4+(a+b)x^3y+(ab+2)x^2y^2+(a+b)axy^3+y^4$$

Equating coefficients, we obtain the system:

$$a+b=0\implies a=-b$$

$$ab+2=1\implies ab=-1$$

And so we find:

$$(a,b)=(\pm1,\mp1)$$

And we may write:

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

This is what we would find by using the difference of squares first, and then applying the sum/difference of cubes, so that's the simpler route to take. :D
 
MarkFL said:
Once you've done that' you've applied the difference of squares...to continue factoring further, as you noted, you need to use the sum/difference of cubes. Now, if you begin with the difference of cubes, you'd have:

$$x^6-y^6=(x^2-y^2)(x^4+x^2y^2+y^4)$$

Now to continue, you would use the difference of squares on the first factor. To factor the second factor, we can try:

$$x^4+x^2y^2+y^4=(x^2+axy+y^2)(x^2+bxy+y^2)=x^4+(a+b)x^3y+(ab+2)x^2y^2+(a+b)axy^3+y^4$$

Equating coefficients, we obtain the system:

$$a+b=0\implies a=-b$$

$$ab+2=1\implies ab=-1$$

And so we find:

$$(a,b)=(\pm1,\mp1)$$

And we may write:

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

This is what we would find by using the difference of squares first, and then applying the sum/difference of cubes, so that's the simpler route to take. :D

I can use u and v in terms of (x^4+x^2y^2+y^4) just like I did in the other post, right?
 
RTCNTC said:
I can use u and v in terms of (x^4+x^2y^2+y^4) just like I did in the other post, right?

You'd wind up with radicals if you do that. :D
 
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