# 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

$$\frac {1}{\sqrt [3]{3} - \sqrt [3]{2}} \label {identity}$$

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

$$\frac {1}{a-b}$$

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

$$a-b = \frac {a^3 - b^3}{a^2 + ab + b^2}$$

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 $$a=\sqrt [3]{3}$$ and $$b=\sqrt [3]{2}$$, 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

$$\frac {1}{\sqrt [3]{3}-\sqrt [3]{2}} = \frac {(\sqrt [3]{3})^2 + \sqrt [3]{3}\sqrt [3]{2} + (\sqrt [3]{2})^2}{1}$$