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!

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 $a=\frac{6}{5}$.
