1. ## factorization

$\displaystyle x^6-y^6$

$\displaystyle x^3y^3(x^3-y^3)$

$\displaystyle x^3y^3(x-y)(x^2+xy+y^2)$

correct? thank u

2. Hello jvignacio:

No, your result is not correct. (You could determine this yourself if you were to multiply it out.)

Rewrite the original expression as a difference of squares.

(x^3)^2 - (y^3)^2

Cheers,

~ Mark

3. Originally Posted by mmm4444bot
Hello jvignacio:

No, your result is not correct. (You could determine this yourself if you were to multiply it out.)

Rewrite the original expression as a difference of squares.

(x^3)^2 - (y^3)^2

Cheers,

~ Mark
so what your saying is $\displaystyle (x^3-y^3)^2$ ??

then it would be

$\displaystyle ((x-y)(x^2+xy+y^2))((x-y)(x^2+xy+y^2))$

4. Originally Posted by jvignacio
so what your saying is $\displaystyle (x^3-y^3)^2$ ??
No, this is not what I said.

A^2 - B^2 is not the same as (A - B)^2.

Do you know how to multiply out these results of yours? If you make an effort to do so, then you will discover for yourself that your results are incorrect.

For example, expand (x^3 - y^3)^2. It equals the following.

x^6 - 2x^3 * y^3 + y^6

Clearly wrong.

Do you know the factorization pattern for a difference of squares?

A^2 - B^2 = (A + B)*(A - B)

Therefore, (A^3)^2 - (B^3)^2 equals the following.

(A^3 + B^3)*(A^3 - B^3)

Both of these factors have well-known factorizations.

The first is a sum of cubes, and the second is a difference of cubes.

Do you see the strategy now?

Cheers,

~ Mark

5. Originally Posted by mmm4444bot
No, this is not what I said.

A^2 - B^2 is not the same as (A - B)^2.

Do you know how to multiply out these results of yours? If you make an effort to do so, then you will discover for yourself that your results are incorrect.

For example, expand (x^3 - y^3)^2. It equals the following.

x^6 - 2x^3 * y^3 + y^6

Clearly wrong.

Do you know the factorization pattern for a difference of squares?

A^2 - B^2 = (A + B)*(A - B)

Therefore, (A^3)^2 - (B^3)^2 equals the following.

(A^3 + B^3)*(A^3 - B^3)

Both of these factors have well-known factorizations.

The first is a sum of cubes, and the second is a difference of cubes.

Do you see the strategy now?

Cheers,

~ Mark
i think i understand now..

its

$\displaystyle (x^3)^2-(y^3)^2$

$\displaystyle = (x^3-y^3)(x^3+y^3)$

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