Question

How do you factor `u^3 - 1`?

Answer 1

`u^3-1 = (u-1)(u^2+u+1)`

Explanation:

There is a useful 'difference of cubes' identity:

`a^3-b^3 = (a-b)(a^2+ab+b^2)`

In our example, `a = u` and `b = 1` :

`u^3-1 = u^3-1^3 = (u-1)(u^2+u*1+1^2) = (u-1)(u^2+u+1)`

In fact, in general we find:

`a^n - b^n = (a-b)(a^(n-1) + a^(n-2)b + a^(n-3)b^2 +...+ b^(n-1))`

When you multiply it out, most of the terms cancel.

When `n` is even, the second factor can be factorised further:

`a^(n-1) + a^(n-2)b + a^(n-3)b^2 +...+ b^(n-1)`

`= (a+b)(a^(n-2) + a^(n-4)b^2 + ... + b^(n-2))`

There are no more linear factors with Real coefficients, but there are quadratic factors with Real coefficients.

For example:

`a^6 - b^6 = (a-b)(a+b)(a^2-ab+b^2)(a^2+ab+b^2)`