How do we find the coefficient of $x^0$ in this expression? $$\left(\frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}}\right)^{10}$$
This problem is just very perplexing. I wrote down the $T_{r+1}$ for it as such:
$T_{r+1} = (-1)^r\;^nC_r\;(\frac{x+1}{x^{2/3}-x^{1/3}+1})^{n-r}\;(\frac{x-1}{x-x^{1/2}})^r$
I tried plugging the expression into Wolfram Alpha, it showed me an error. I tried a few other sites, but they were all in vain.
Ordinarily, there should be only one unknown term having $x$, whose power we equate to $0$, but that clearly isn't the case here. It should be solved without using Trinomial or Multinomial Theorem. Is there any way of doing so, and finding the coefficient of $x^0$?
Edit: Thanks to the users in the comments, I could identify the $a^3 + b^3$ and $a^2-b^2$ identities. I'm now getting the coefficient of $x^0$ as $210$, which is correct.
$$\left(\frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}}\right)^{10}$$
We can simplify both the terms as follows;
The first term $\left(\frac{x+1}{x^{2/3}-x^{1/3}+1}\right)$, take $x^{\frac{1}{3}}=t$ and it simplifies to $x^{\frac{1}{3}}+1$
Similarly,
The second term $\left(\frac{x-1}{x-x^{1/2}}\right)$, take $x^{\frac{1}{2}}=t$ and it simplifies to $x^{\frac{-1}{2}}+1$
$$(x^{\frac{1}{3}}-x^{\frac{-1}{2}})^{10}$$
Writing the general term;
$$T_{r+1}=\,^nc_r(x^{\frac{10-r}{3}})(-x^{\frac{-r}{2}})$$
$$T_{r+1}=(-1)^r\,^{10}c_r(x^{\frac{10-r}{3}-\frac{r}{2}})$$ $$T_{r+1}=(-1)^r\,^{10}c_r(x^{\frac{20-5r}{6}})$$
To get the coefficient of $x^0$, we can plug in $r=4$;
coefficient of $x^0=\,^{10}c_4=210$