Once you get hold of it, it's easy -Part 2

To make it easier put $x=n+10\Longrightarrow n=x-10$

So $\displaystyle \dfrac{n^{3}+100}{n+10}$ reduces to $\frac { { (x-10) }^{ 3 }+100 }{ x } =\frac { { (x }^{ 3 }-30{ x }^{ 2 }+300x-1000)+100 }{ x }$

$=x^{ 2 }-30x+300-\frac { 900 }{ x }$

The largest divisor of 900 is 900 itself,

Therefore $n=900-10=\boxed{890}$

We can write, $n+10 | n^3+100$

$\Rightarrow n+10| n^2(n+10)-10n(n+10)+100(n+10)-900$

$\Rightarrow n+10|900$

$\Rightarrow (n+10)_{max}=900$

$\Rightarrow n_{max}=890$

(n³ + 100)/(n+10) = (n³ + 1000- 900)/(n+10) = n² -10 n+100 - 900/(n+10)

Then

n +10 is a divisor of 900

The greatest divisor of 900 is 900

Then

to obtain the greatest value of n, we must have n + 10 = 900

So

n = 890

$\frac {n^3 + 100} {n+10} =\frac {n^3 + 1000 - 900} {n+10} =n^2 - 10n + 100 - \frac {900} {n+10}$

For $\frac {n^3 + 100} {n+10}$ to be an integer, $\frac {900} {n+10}$ must also be an integer.

The largest $n$ possible occurs when $\frac {900} {n+10} =1$ , which means $n + 10 = 900$

Hence, $n = \boxed {890}$

n³ + 100 = n³ + 10³ - 900 = (n+10)(n²-10n+100) - 900

So 900 must be divisible by n+10.

The largest divisor of 900 is 900, so n = 890

Let $\frac{n^3 + 100}{n+10} = a + \frac{b}{n+10}$ $\large \ Now \ using \ partial \ fractions:$ $a(n+10) + b = n^3 +100 \Rightarrow\ let \ n = -10 \Rightarrow\ b = -900$ $\large \ rearranging \ what's \ left \ over:$ $a = \frac{n^3 +1000}{n+10} = \frac{(n+10)^3 - 30n(n+10)}{n+10} = (n+10)^2 -30n$ $\Rightarrow\frac{n^3 +100}{n+10} = (n+10)^2 -30n - \frac{900}{n+10} \therefore\ max(n) = \boxed{890}$