Proof that certain number is an integer Let $k$ be an integer and let
$$
n=\sqrt[3]{k+\sqrt{k^2-1}}+\sqrt[3]{k-\sqrt{k^2-1}}+1
$$
Prove that $n^3-3n^2$ is an integer.

(I have started posting any problem I get stuck on and then subsequently find a good solution to here on math.se, primarily to get new solutions which might be even better, and to get checked if my solutions are ok)

Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$a+b+c=0\implies a^3+b^3+c^3=3abc$$
Let $a=\sqrt[3]{k+\sqrt{k^2-1}}$, $b=\sqrt[3]{k-\sqrt{k^2-1}}$, $c=1-n$
Clearly we have $a+b+c=0$
$$a^3+b^3+c^3=3abc\tag{1}$$
$$ab=\sqrt[3]{k+\sqrt{k^2-1}}\cdot\sqrt[3]{k-\sqrt{k^2-1}}$$
$$ab=\sqrt[3]{k^2-(\sqrt{k^2-1})^2}$$
$$ab=\sqrt[3]1 = 1$$
Substituting back in $(1)$
$$k+\sqrt{k^2-1}+k-\sqrt{k^2-1}+(1-n)^3=3(1-n)$$
$$2k+1-3n+3n^2-n^3 = 3-3n$$
$$\require{cancel}{2k+1-\cancel{3n}+3n^2-n^3 = 3-\cancel{3n}}$$
$$n^3-3n^2=2k-2$$
which is obviously an integer.
