Understanding Limits: Solving a Tricky Step in a Limits Problem

[Stevo6754]

Homework Statement

lim$$\frac{x^{1/3}-a^{1/3}}{x - a}$$ as a approaches 0

Homework Equations

The Attempt at a Solution

Heres my problem, I can do limits no problem, but my book shows steps on how to do the problem, there is one particular step I don't understand. It is how they got from the above equation to

lim$$\frac{x^{1/3}-a^{1/3}}{(x^{1/3}-a^{1/3})(x^{2/3}+x^{1/3}a^{1/3}+a^{2/3})}$$ok so if I multiple the denominator out I get x - a, so it seems like they changed the denominator to a polynomial that they could factor out the same value as the numerator yet still = x - a. Am I understanding this right?

edit - sorry I butchered the format some due to previewing it my post 23432 times.

 

[Dick]

Yes, you are understanding it right. x-a=(x^(1/3))^3-(a^(1/3))^3. They factored x-a to to cancel the numerator.

 

[Stevo6754]

Yes, you are understanding it right. x-a=(x^(1/3))^3-(a^(1/3))^3. They factored x-a to to cancel the numerator.

Ok, but how did they come up with that polynomial? They needed to find a value they could multiply the numerator by to get the value in the denominator? Is there some fundamental thing I should know about how they came up with this polynomial?

 

[Dick]

a^3-b^3=(a-b)*(a^2+ba+b^2). They just expressed x-a as a difference of cubes, which were, of course, the cubes of x^(1/3) and a^(1/3). It's not that fundamental. It's just an algebraic trick.

 

[Stevo6754]

a^3-b^3=(a-b)*(a^2+ba+b^2). They just expressed x-a as a difference of cubes, which were, of course, the cubes of x^(1/3) and a^(1/3). It's not that fundamental. It's just an algebraic trick.

Great! Thanks Dick.

 

[Dick]

Ok, but how did they come up with that polynomial? They needed to find a value they could multiply the numerator by to get the value in the denominator? Is there some fundamental thing I should know about how they came up with this polynomial?

They just used that (a^3-b^3)=(a-b)*(a^2+ab+b^2). There are similar factorizations for (a^n-b^n) for any n. It's just an algebraic trick.