NT is quite popular

So we want to find largest integer $n$ such that $\dfrac{n^3+100}{n+10}$ is an integer,

$\dfrac{n^3+100}{n+10} = \dfrac{n^3+1000-900}{n+10}=\dfrac{(n+10)(n^2-10n+100)-900}{n+10} =n^2-10n+100 -\dfrac{900}{n+10}$

For the above expression to be an integer , the largest value of $n$ is $890$ .

Let,

$\begin{aligned} n^3+100 & =(n^2+an+b)(n+10)+c \\ & = n^3+n^2(10+a)+n(b+10a)+10b+c \\ \end{aligned} \\ \begin{cases} 10+a &= 0 \Rightarrow a=-10 \\ b+10a & = 0 \Rightarrow b=100 \\ c+10b & = 100 \Rightarrow c=-900 \end{cases} \\ \gcd(n^3+100,n+10)=\gcd(-900,n+10)=\gcd(900,n+10) \\ \text{Maximum value of }n=\boxed{890}$