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 $x=1+\sqrt{\frac{152}{27}},~~y=-1+\sqrt{\frac{152}{27}}$
$$x^{\frac{1}3}y^{\frac{1}3}=\frac{5}{3}$$
$$a=x^{\frac{1}3}-y^{\frac{1}3}$$
$$\begin{align}
a\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)&=\left(x^{\frac{1}3}-y^{\frac{1}3}\right)\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)\\
\\
a\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)&=x-y=2\\
\\
a\left(\left(x^{\frac{1}3}-y^{\frac{1}3}\right)^2+3x^{\frac{1}3}y^{\frac{1}3}\right)&=2\\
\\
a\left(a^2+5\right)&=2\\
\\
a^3+5a&=2
\end{align}$$
