Difference of Cubes?

Question

I have a real hard time understading the difference of cubes.
However, there are several questions on this topic in the coming June state exam that I must prepare for.

How do I play with this topic?

QUESTIONS:

(1) Factor a^3b^3 - 8x^6y^9

(2) x^3 + 27

Answer 1

here is the formula
(a^3 - b^3) = (a-b)(a^2+ab+b^2)

1) a^3b^3 - 8x^6y^9
a^3b^3 - 2^3(x^2)^3(y^3)^3
(ab)^3 - (2x^2y^3)^3
(ab - 2x^2y^3)(a^2b^2 + 2abx^2y^3 - 4x^4y^6)

2)(a^3 + b^3) = (a+b)(a^2 -ab +b^2)
x^3 + 27
x^3 + 3^3
(x+3)(x^2 - 3x + 9)

Answer 2

The thing to remember with difference of cubes is that (a-b) is a factor. You can work out the quadratic part from there using polynomial short division:
(a^3 - b^3) = (a-b) (a^2 + ab + b^2)
Then you can apply this to more complicated things.

Note that the second factor above is (a + b/2)^2 + 3b^2/4; as the sum of squares it can only be zero if both (a + b/2) and b^2 are 0, i.e. a = b = 0. So it is irreducible.

Sum of cubes is basically the same as difference of cubes, just replace b with -b (since a^3 - b^3 = a^3 + (-b)^3).

e.g.
(1) a^3b^3 - 8x^6y^9
= (ab)^3 - (2x^2y^3)^3
= (ab - 2x^2y^3) ((ab)^2 + (ab)(2x^2y^3) + (2x^2y^3)^2)
= (ab - 2x^2y^3) (a^2b^2 + 2abx^2y^3 + 4x^4y^6)

(2) x^3 + 27
= (x+3) (x^2 - 3x + 9)