<s>[INST] Finding the number of solutions of an expression Find the number of integers $1\leq x \leq 2010 $ such that the expression 
$\sqrt[3]{x+(x+8)\sqrt\frac{x-1}{27}}-\sqrt[3]{x-(x+8)\sqrt\frac{x-1}{27}}$
is a rational number 
NOTE -> I LITERALLY do not understand where to start !! HELP ME!!
 [/INST] The number of integers $1≤x≤2010$, (such that the expression above is a rational number), is 26.
The value of  $x$ is
$x=3n^{2}-6n+4$.
In this way,
$\sqrt{\frac{x-1}{27}}=\sqrt{\frac{3n^{2}-6n+3}{27}}=\sqrt{\frac{3(n-1)^{2}}{27}}=\frac{n-1}{3}$;
$(x+8)\sqrt{\frac{x-1}{27}}=(n^{2}-2n+4)(n-1)$;
$x+(x+8)\sqrt{\frac{x-1}{27}}=(3n^{2}-6n+4)+(n^{2}-2n+4)(n-1)=n^{3}$;
$x-(x+8)\sqrt{\frac{x-1}{27}}=(3n^{2}-6n+4)-(n^{2}-2n+4)(n-1)=(2-n)^{3}$;
$\sqrt[3]{x+(x+8)\sqrt\frac{x-1}{27}}-
\sqrt[3]{x-(x+8)\sqrt\frac{x-1}{27}}=$
$=n-(2-n)=2(n-1)$.
The number $n$ such as $1≤x≤2010$, is the integer solution of
$1≤3n^{2}-6n+4≤2010$,
that is 26.
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