# A calculus of the absurd

##### C.0.3 Rationalising
• Example C.0.2 Show that $$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$, and hence rationalise the denominator of

\begin{equation} \frac {1}{\sqrt {3} - \sqrt {2}} \label {identity} \end{equation}

Solution

The first part is just algebraic manipulation, i.e.

\begin{align} (a-b)(a^2 + ab + b^2) &= a^2 + a^2b + ab^2 - ba^2 - ab^2 - b^3 \\ &= a^2 + (a^2b - ab^2) + (ab^2 - ab^2) - b^3 \\ &= a^3 - b^3 \end{align}

For the second part, we want to apply this identity to the expression to rationalise. We have an expression in the form

\begin{equation} \frac {1}{a-b} \end{equation}

And using the identity (Equation C.20), we know that

\begin{equation} a-b = \frac {a^3 - b^3}{a^2 + ab + b^2} \end{equation}

Which implies (using the properties of reciprocals and fractions150150 See Section 4.1.1 if you’re unsure about this) that

\begin{align} \frac {1}{a-b} &= \frac {a^2 + ab + b^2}{a^3 - b^3} \end{align}

Given that our $$a=\sqrt {3}$$ and $$b=\sqrt {2}$$, this is exactly what we want! As $$a^3=3$$ (which is rational) and $$b^3=2$$ which is rational, $$a^3 - b^3$$ is also rational (specifically $$a-b=1$$). Therefore, we have that

\begin{equation} \frac {1}{\sqrt {3}-\sqrt {2}} = \frac {(\sqrt {3})^2 + \sqrt {3}\sqrt {2} + (\sqrt {2})^2}{1} \end{equation}