$$\epsilon^\frac32=\mu^\frac32+\frac{\left(\pi k_B T\right)^2}{8}\frac1{\mu^\frac12}$$ I am given that $\epsilon\approx\mu$ and $(1+x)^\alpha\approx1+\alpha x$ when $x\to0$.
I have tried manipulating this to $$1-\frac{8}{(\pi k_B T)^2}\mu^\frac12(\epsilon^\frac32-\mu^\frac32)=0\Leftrightarrow\left[1+\mu^\frac12(\epsilon^\frac32-\mu^\frac32)\right]^{-\frac{8}{(\pi k_B T)^2}}=0$$ but this doesn't seem to lead anywhere. any hints please?
the solution is one of these $T_f=\epsilon/k_B$:
$$\epsilon^\frac32=\mu^\frac32+\frac{\left(\pi k_B T\right)^2}{8}\frac1{\mu^\frac12}$$
Let $b=\frac{\left(\pi k_B T\right)^2}{8}$ and croos mulitiply to make $$f(\mu)=\epsilon ^{3/2}\sqrt{\mu }-\mu ^2-b=0$$ Now, since you are told that $\epsilon\approx\mu$, expand $f(\mu)$ as a series around $\mu=\epsilon$. This gives $$f(\mu)=-b-\frac{3}{2} \epsilon (\mu -\epsilon )+O\left((\mu -\epsilon )^2\right)$$ Neglecting the higher order terms $$\mu=\epsilon -\frac{2 b}{3 \epsilon }=\epsilon \left(1-\frac{2 b}{3 \epsilon ^2}\right)$$ Just continue and introduce $T_f$.