Factorising a Cubic: An Easy Way?

[tmonk]

I am attempting to derive the formula for the variance of a continuous uniform distribution where f(x) = \left\{ \begin{array}{ll}<br /> \frac{1}{b-a} &amp; \mbox{$a \leq $x$ \leq b$};\\<br /> 0 &amp; \mbox{otherwise}.\end{array} \right.

I was successful, but only after using Wolfram Alpha to factorise the cubic.

By using Var[x] = \int_a^b \! x^{2}f(x)\, \mathrm{d}x - \mu^{2}
I got this:\frac{b^{3} - a^{3} + 3a^{2}b - 3ab^{2}}{12(b-a)}
The numerator just factorises to give (b-a)³, which gives the correct formula of \frac {(b-a)^{2}}{12}

I tried using algebraic division to divide by the common factor of b-a, but I don't think I did it right. Is there an easy way to factorise the cubic?

 

[micromass]

You should remember the formula

(a+b)^n=\sum_{k=0}^n{\binom{n}{k}a^kb^{n-k}}

well. It will come in handy a lot of time...

That said, it's always possible to factorize a cubic, but it's not always easy. See http://en.wikipedia.org/wiki/Cubic_equation

 

[I like Serena]

I tried using algebraic division to divide by the common factor of b-a, but I don't think I did it right. Is there an easy way to factorise the cubic?

Just for fun.
Here's the method to do algebraic division on polynomials.

b-a | b[SUP]3[/SUP] - 3ab[SUP]2[/SUP] + 3a[SUP]2[/SUP]b - a[SUP]3[/SUP] | b[SUP]2[/SUP] ...
      b[SUP]3[/SUP] -  ab[SUP]2[/SUP]
      ----------
          -2ab[SUP]2[/SUP] + 3a[SUP]2[/SUP]b
          ...

@MM: Hey ;)