And we thought binomial is harmless

Using the identity $a^3 + b^3 = (a+b)(a^2+b^2-ab)$ , we have $\frac{y+1}{y^{\frac{2}{3}} - y^{\frac{1}{3}} + 1 } = y^{\frac{1}{3}} + 1$ Also, using the identity $a^2-b^2 =(a-b)(a+b)$ , we have $\frac{y-1}{y-y^{\frac{1}{2}}} = y^{-\frac{1}{2}} + 1$ Thus, all we have to do is find the constant term in the expansion of $(y^{\frac{1}{3}} - y^{-\frac{1}{2}})^{10}$ . From binomal theorm, we have the $r^{\textrm{th}}$ term as $\binom{10}{r} (-1)^{10-r} y^{\frac{r}{3} - \frac{10-r}{2}}$ . Since for a constant term, the power of $y$ is zero, we must have $\frac{r}{3} -\frac{10-r}{2} = 0 \Rightarrow r = 6$ Thus, the constant term is $\binom{10}{6} = \boxed{210}$