How can the numerator of a rational expression with cube roots be simplified?

[mindauggas]

Homework Statement

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

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?

 

[Mark44]

Homework Statement

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

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?

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

 

[mindauggas]

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