1. Special Factoring

$\displaystyle (a - b)^3 - b^3$
$\displaystyle [(a - b) - b][(a - b)^2 + (a - b)b + b^2]$

Is that right?

. . . Nice going!

3. Gah, this one is driving me nuts :

Factor:

$\displaystyle 27a^4 - a$
$\displaystyle a(27a^3 - 1)$

Or can I do more?

Gah, this one is driving me nuts :

Factor:

$\displaystyle 27a^4 - a$
$\displaystyle a(27a^3 - 1)$

Or can I do more?
Note that $\displaystyle 27a^3 - 1 = (3a)^3 - 1^3$, a difference of two perfect cubes ......

5. I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks.

EDIT: Just in case...

$\displaystyle (a)(27a - 1)(27a^2 + 27a + 1)$

I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks.

EDIT: Just in case...

$\displaystyle (a)(27a - 1)(27a^2 + 27a + 1)$
Should be 3a's, NOT 27a's ......

7. Sorry, I don't understand. I looked back over my notes but I'm kind of confused.

Gah, this one is driving me nuts :

Factor:

$\displaystyle 27a^4 - a$
$\displaystyle a(27a^3 - 1)$

Or can I do more?
$\displaystyle 27a^3 - 1 = (3a)^3 - (1)^3$

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

So let $\displaystyle x = 3a$ and $\displaystyle y = 1$

Thus
$\displaystyle 27a^3 - 1 = (3a)^3 - (1)^3 = (3a - 1)((3a)^2 + (3a)(1) + (1)^2)$

$\displaystyle = (3a - 1)(9a^2 + 3a + 1)$

-Dan

9. Oh, now I understand, thank you.