X^6 - y^6 As Difference of Squares

[mathdad]

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?

 

[MarkFL]

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:

\(\displaystyle x^6-y^6=(x^2-y^2)(x^4+x^2y^2+y^4)\)

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

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

Equating coefficients, we obtain the system:

\(\displaystyle a+b=0\implies a=-b\)

\(\displaystyle ab+2=1\implies ab=-1\)

And so we find:

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

And we may write:

\(\displaystyle x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)\)

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

 

[mathdad]

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:

\(\displaystyle x^6-y^6=(x^2-y^2)(x^4+x^2y^2+y^4)\)

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

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

Equating coefficients, we obtain the system:

\(\displaystyle a+b=0\implies a=-b\)

\(\displaystyle ab+2=1\implies ab=-1\)

And so we find:

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

And we may write:

\(\displaystyle x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)\)

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?

 

[MarkFL]

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

 

[mathdad]

Good information.