# Homework Help: Rationalise the numerator

1. Nov 26, 2011

### mindauggas

1. The problem statement, all variables and given/known data

$$\frac{\sqrt[3]{x}-\sqrt[3]{a}}{x-a}$$

3. The attempt at a solution

$\frac{(\sqrt[3]{x}-\sqrt[3]{a})*(\sqrt[3]{x}+\sqrt[3]{a})}{(x-a)(\sqrt[3]{x}+\sqrt[3]{a})}$

We get:

$\frac{\sqrt[3]{x^{2}}-\sqrt[3]{a^{2}}}{\sqrt[3]{x^{4}}+x\sqrt[3]{a}-a\sqrt[3]{x}-\sqrt[3]{a^{4}}}$

Don't know what to do next. Am I even on the right track or should I multiply the numerator and the denominator with smth different?

Last edited: Nov 26, 2011
2. Nov 26, 2011

### Staff: Mentor

Something different. If you have a difference of square roots, such as √x - √y, you would multiply by √x + √y over itself. This uses the idea that (a - b)(a + b) = a2 - b2, so if a and b are square roots, squaring them gets rid of the radicals.

If you have a difference of cube roots, as in your problem, multiplying by the sum of cube roots doesn't get rid of the radicals. What you need to do is take advantage of this formula:
(a - b)(a2 + ab + b2) = a3 - b3

Notice that if a and b are cube roots, the final expression won't involve cube roots.

There's a similar formula for the sum of cubes:
(a + b)(a2 - ab + b2) = a3 + b3

3. Nov 26, 2011

### mindauggas

Seem's so easy when you know how to do it ... thank you.