How do you factor $u^{3} - 1$ $u^3 - 1$ ?

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Answer 1 · 2015-09-23T18:08:42

$u^{3} - 1 = (u - 1) (u^{2} + u + 1)$ $u^3-1 = (u-1)(u^2+u+1)$

There is a useful 'difference of cubes' identity:

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3 = (a-b)(a^2+ab+b^2)$

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

$u^{3} - 1 = u^{3} - 1^{3} = (u - 1) (u^{2} + u \cdot 1 + 1^{2}) = (u - 1) (u^{2} + u + 1)$ $u^3-1 = u^3-1^3 = (u-1)(u^2+u*1+1^2) = (u-1)(u^2+u+1)$

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:

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

How do you factor u^3 - 1u3−1u^3 - 1?