Finding the number of solutions of an expression

Question

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!!

Answer 1

Hint: Define $$a=\sqrt[3]{x+(x+8)\sqrt\frac{x-1}{27}},$$ $$b=\sqrt[3]{x-(x+8)\sqrt\frac{x-1}{27}},$$ $$u = a-b$$

Determine $u^3=a^3-3a^2b+3ab^2-b^3=a^3-3ab(a-b)-b^3=a^3-3abu-b^3$. Note the simplyfications. Can you complete it from here. Btw you could also brute force this problem with Excel :D

Answer 2

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.