Find a, real number for the following limit Find a, real number, such that $\displaystyle{\lim_{n\to\infty}{(n^3+an^2)^\frac{1}{3}-(n^2-an)^\frac{1}{2}}}=1$
I noted x=$(n^3+an^2)^\frac{1}{3}$
and y=$(n^2-an)^\frac{1}{2}$
I applied with the conjugate ($x^2-xy+y^2$) but I do not know how to continue. Any ideas?
Thank you for your help!
 A: Note that
$$\begin{align*}
x-y &= (x-n) + (n-y) \\
&= \frac{x^3-n^3}{x^2+xn+n^2} + \frac{n^2-y^2}{n+y} \\
&= \frac{an^2}{x^2+xn+n^2} + \frac{an}{n+y} \\
&= \frac{a}{(x/n)^2 + (x/n) + 1} + \frac{a}{1+(y/n)}\end{align*}$$
and therefore, as $\lim_{n\to\infty} \frac{x}{n} = \lim_{n\to\infty} \frac{y}{n} = 1$, $$\lim_{n\to\infty} (x-y) = \lim_{n\to \infty} \left(\frac{a}{(x/n)^2 + (x/n) + 1} + \frac{a}{1+(y/n)}\right) = \frac{a}{1+1+1}+\frac{a}{1+1} = \frac{5a}{6}$$
Setting $\frac{5a}{6} = 1$ yields $a=\frac{6}{5}$.
A: Let $\alpha=n^3+an^2$ and $\beta=n^2-an$. Introduce a difference of $6^{\rm th}$ powers:
$$\begin{align*}
\alpha^{1/3} - \beta^{1/2} &= (\alpha^2)^{1/6} - (\beta^3)^{1/6} \\[1ex]
&= \frac{\left((\alpha^2)^{1/6}\right)^6 - \left((\beta^3)^{1/6}\right)^6}{(\alpha^2)^{5/6} + (\alpha^2)^{4/6}(\beta^3)^{1/6} + (\alpha^2)^{3/6} (\beta^3)^{2/6} + (\alpha^2)^{2/6} (\beta^3)^{3/6} + (\alpha^2)^{1/6} (\beta^3)^{4/6} + (\beta^3)^{5/6}} \\[1ex]
&= \frac{\alpha^2 - \beta^3}{\alpha^{5/3} + \alpha^{4/3} \beta^{1/2} + \alpha \beta + \alpha^{2/3} \beta^{3/2} + \alpha^{1/3} \beta^2 + \beta^{5/2}} \\[1ex]
&= \frac{5an^5 - 2a^2 n^4 + a^3 n^3}{(1+\gamma_+^{5/3}+\gamma_+^{4/3}\gamma_-^{1/2}+\gamma_+^{2/3}\gamma_-^{3/2} + \gamma_+^{1/3}\gamma_-^2 + \gamma_-^{5/2})n^5 - an^4 + an^3 - a^2n^2}
\end{align*}$$
where $\gamma_{\pm}=1\pm\frac an$. Both $\gamma_{\pm}\to1$ as $n\to\infty$, so the overall expression converges to $\frac{5a}{6}$, hence $a=\frac65$.
