If $n$ is an integer such that $\frac{|n+6|-|n-3|}{\sqrt{100-n^3}} \geq 0$, then the number of possible values of $n$ is ___

For the expression to be defined, the denominator $\sqrt{100 - n^3}$ must be a positive real number, so:

$100 - n^3 > 0 \implies n^3 < 100$

The largest integer satisfying this is $n = 4$ (since $4^3 = 64 < 100$ but $5^3 = 125 > 100$).

So $n \leq 4$.

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Since the denominator is positive, the inequality reduces to:

$|n + 6| - |n - 3| \geq 0$

$|n + 6| \geq |n - 3|$

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Both sides are non-negative, so squaring is safe:

$(n + 6)^2 \geq (n - 3)^2$

$n^2 + 12n + 36 \geq n^2 - 6n + 9$

$18n \geq -27$

$n \geq -\dfrac{3}{2}$

So $n \geq -1$ (smallest integer satisfying this).

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Combining $-1 \leq n \leq 4$, the integer values are $\{-1, 0, 1, 2, 3, 4\}$.

Number of possible values $= 6$.