1. ## Special Factoring Formulas

Does anyone know how to derive the factor formulas for sum and difference of cubes? that is
$\displaystyle a^3 + b^3 = (a + b) (a^2 - ab + b^2)$
$\displaystyle a^3 - b^3 = (a - b) (a^2 + ab + b^2)$

2. Originally Posted by skylabel
Does anyone know how to derive the factor formulas for sum and difference of cubes? that is
$\displaystyle a^3 + b^3 = (a + b) (a^2 - ab + b^2)$
$\displaystyle a^3 - b^3 = (a - b) (a^2 + ab + b^2)$
$\displaystyle a^3 + b^3$

$\displaystyle a^3 -a^2b+ab^2+ ba^2 - ab^2 + b^3$

$\displaystyle a^3+ ba^2 - a^2b - ab^2 +ab^2+ b^3$

$\displaystyle a^2(a+b) - ab(a+b) +b^2(a+b)$

$\displaystyle (a^2 - ab + b^2)(a+b)$
.................................................. ...........................
$\displaystyle a^3 - b^3$

$\displaystyle a^3 -a^2b+a^2b+ b^2a - ab^2 - b^3$

$\displaystyle a^3- ba^2 + a^2b - ab^2 +ab^2- b^3$

$\displaystyle a^2(a+b) + ab(a-b) +b^2(a-b)$

$\displaystyle (a^2 + ab + b^2)(a-b)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I have just multiplied the factors to show you the steps

3. Originally Posted by skylabel
Does anyone know how to derive the factor formulas for sum and difference of cubes?
Another method is to do the polynomial long division:

[HTML] a^2 + ab + b^2
--------------------------
a - b )a^3 + 0a^2b + 0ab^2 - b^3
a^3 - 1a^2b
--------------------------
1a^2b + 0ab^2 - b^3
1a^2b - 1ab^2
--------------------------
1ab^2 - b^3
1ab^2 - b^3
--------------------------
0[/HTML]

4. Originally Posted by stapel
Another method is to do the polynomial long division:

[html] a^2 + ab + b^2
--------------------------
a - b )a^3 + 0a^2b + 0ab^2 - b^3
a^3 - 1a^2b
--------------------------
1a^2b + 0ab^2 - b^3
1a^2b - 1ab^2
--------------------------
1ab^2 - b^3
1ab^2 - b^3
--------------------------
0[/html]
To elaborate:

Since a^3 - b^3 is equal to zero when a = b, it's obvious that a - b is a factor of a^3 - b^3.