Show that $a^3+5a$ is an integer I've been given the following task. Let
$$a = \sqrt[3]{1+\sqrt{\frac{152}{27}}}-\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$$
Show that $a^3+5a$ is an integer.
I tried calculating it by hand but the small page of my copybook is not large enough for the very long calculations. Is there a trick I could use here instead of calculating it by hand?
I tried factoring $a^3+5a$ into $a(a^2+5)$ to make it more simple to calculate but it still gets a tad bit complicated when multiplying with $a$ again.
Thank you in advance.

Let $x=\sqrt[3]{1+\sqrt{\frac{152}{27}}}$ and $y=\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$ .
Then
$$\begin{aligned}a^3+5a&=(x-y)^3+5(x-y)\\
&=x^3-3xy(x-y)-y^3+5(x-y)\\
&=x^3-y^3+(5-3xy)(x-y)
\end{aligned}$$
Now,
$\begin{aligned}5-3xy&=5-3\sqrt[3]{\left(1+\sqrt{\frac{152}{27}}\right)\left(-1+\sqrt{\frac{152}{27}}\right)}\\
&=5-3\sqrt[3]{\frac{125}{27}}\\
&=5-3\cdot\frac53\\
&=0
\end{aligned}$
and $x^3-y^3=2$, so $a^3+5a=2$.
