Question

Is there any short way to expand \(\dfrac{x^{1/3}-a^{1/3}}{x-a}\)
Please help

Answer 1

@SithsAndGiggles

Answer 2

\(\large \begin{align} \color{black}{
\normalsize \text{let } x^{\frac{1}{3}}=m \normalsize \text{and let } a^{\frac{1}{3}}=n \hspace{.33em}\\~\\
\dfrac{x^{\frac{1}{3}}-a^{\frac{1}{3}}}{x-a} \hspace{.33em}\\~\\
=\dfrac{m-n}{m^3-n^3} \hspace{.33em}\\~\\
=\dfrac{1}{m^2+mn+n^2} \hspace{.33em}\\~\\
=\dfrac{1}{x^{\frac{2}{3}}+x^{\frac{1}{3}}a^{\frac{1}{3}}+a^{\frac{2}{3}}} \hspace{.33em}\\~\\
}\end{align}\)


see if that are asking about

Answer 3

Thank you so much

Answer 4

was easy

Answer 5

Yes, for you only, not for me. I thought of polynomial expansion and got stuck. You can see what wolfram does. OMG, no way !! http://www.wolframalpha.com/input/?i=expand+%28x^%281%2F3+%29-a^%281%2F3%29%29%2F%28%28x-a%29

Answer 6

http://www.wolframalpha.com/input/?i=simplify+%28x%5E%281%2F3+%29-a%5E%281%2F3%29%29%2F%28%28x-a%29

Answer 7

ur link is showing some series expansion

Answer 8

Nope, My Prof asked me to show work. hehehe...

Answer 9

Since it is part of the problem. I need show f(x) = x^(1/3) differentiable at x =a (a is not 0) and that f'(a) = 1/(3a^(2/3)