<s>[INST] Simplify the following expression! 
Simplify the following expression

$$\sqrt[3]{a+\frac{a+8}{3}\sqrt{\frac{a-1}{3}}}+\sqrt[3]{a-\frac{a+8}{3}\sqrt{\frac{a-1}{3}}}$$
I tried using the form $\displaystyle\frac{a^3+b^3}{a^2-ab+b^2}$, and i also tried to assume the requested expression with $c$ and use the form $a^3+b^3-c^3=-3abc$, if $a+b-c= 0$, but haven't obtained its simple form yet,
i get this problem from a book but i doesn't get the solution

if anyone can complete it it's really amazing

 [/INST] This answer was started some longer time ago, now i finally had time to complete it. Although we already have a good, compact, accepted answer i will still post it since the path of getting the solution is (slightly) different, and may be of interest.

Let $u,v$ be the two numbers
$$
\sqrt[3]{a\pm\frac {a+8}3\sqrt{\frac{a-1}3}}\ .
$$
Then we have $u^3+v^3=2a$ and
$$
\begin{aligned}
(uv)^3 &=u^3v^3
\\
&=
\left(a+\frac {a+8}3\sqrt{\frac{a-1}3}\right)
\left(a-\frac {a+8}3\sqrt{\frac{a-1}3}\right)
\\
&=\frac 1{27}\left(27a^2-(a+8)^2(a-1)\right)
\\
&=-\frac 1{27}(a-4)^3
\ .
\end{aligned}
$$
Assuming that $u,v\in\Bbb R$, and this happens in the case and only in the case of $\sqrt{(a-1)/3}\in\Bbb R$, i.e. $a\ge 1$, (and only when we agree to take two real values in $\sqrt[3]{\dots}$) we obtain $$uv=-\frac 13(a-4)\ .$$
This implies:
$$
\begin{aligned}
(u+v)^3 &= (u^3+v^3)+3uv(u+v)\\
&=2a - (a-4)(u+v)\ ,
\end{aligned}
$$
so $u+v$ is a root of the polynomial of degree three $X^3 + (a-4)X -2a=(X-2)(X^2+2X+a)$.

*

*For $a=1$ we easily check $u=v=1$, so $u+v=2$. Else...

*For $a>1$ we can (and as assumed do) perform all computations with $u,v\in\Bbb R$, so $(u+v)$ is the only real solution of $(X-2)(X^2+2X+a)$, which is $2$.

*For $a<1$ or for a general complex $a$ we still can build $u,v\in\Bbb C$. Assuming that we take the two roots $\sqrt[3]{\dots}$ in $u,v$ in $\Bbb C$ so that we still have the compatibility $uv=-(a-4)/3$ and we obtain at any rate a root of the mentioned polynomial for the value $u+v$.

</s>