Finding Binomial Expansion as x approaches infinity of a function I am trying to find the series that is valid for
$(x^3 + x^2 + 1)^{\frac{1}{3}} - (x^2 + x)^{\frac{1}{2}}$
as x approaches infinity.
The question says this should be possible using the binomial expansion but I can't work out how to do that.

$$(x^3 + x^2 + 1)^{\frac{1}{3}} - (x^2 + x)^{\frac{1}{2}}=x\left(1+\frac{x^2+1}{x^3}\right)^{\frac{1}{3}}-x  \left(1+\frac{1}{x}\right)^{\frac{1}{2}}$$
Now use
$$(1+\epsilon )^{\frac{1}{n}}=\sum_{n=0}^\infty \binom{\frac{1}{n}}{n}\epsilon^n$$
Fot the first term, make $n=\frac 13$ and $\epsilon= \frac{x^2+1}{x^3}$; for the second , make $n=\frac 12$ and $\epsilon= \frac{1}{x}$.
Expand a little the powers of $\frac{x^2+1}{x^3}$ and you will have
$$(x^3 + x^2 + 1)^{\frac{1}{3}} - (x^2 + x)^{\frac{1}{2}}=-\frac{1}{6}+\frac{1}{72 x}+\frac{431}{1296 x^2}-\frac{6977}{31104
   x^3}+O\left(\frac{1}{x^4}\right)$$
Try for $x=1000$; the truncated series will give
$$-\frac{5183557662977}{31104000000000} =\color{red}{-0.166652445440}361$$ while
$$\sqrt[3]{1001000001}-10 \sqrt{10010}=\color{red}{-0.166652445440174}$$
