Solution for $n^{}$

n (n^2 -1) = 24 n; n^2 - 25 = 0; (n - 5)(n + 5) = 0; The solution for n is +5 and -5.

Since $n$ is a common factor on both sides, $n$ cancels out. So we have

$n^2-1=24$

$n^2=25$

$n=\pm 5$

The desired answer is $5$ .

n(n^2-1)=24n; divide n in both sides n^2-1=24; move 24 to left side n^2-25=0; (n+5)(n-5)=0 n=5 or n=-5; since n is a positive integer, n=5

$\color{#3D99F6}{n(n^2-1)=24n\rightarrow=n^3-n=24n}$ $\color{#D61F06}{n^3-25n=0\rightarrow n(n+5)(n-5)=0}$ $\color{#69047E}{n=0,5\;or\;-5}$ As $n$ is a positive integer so we discard the negative root and we have $\color{#EC7300}{n=\boxed{5}}$

$n^{3}$ - $n^{}$ = 24 $n^{}$
$n^{3}$ = 25 $n^{}$
$\frac{n^3}{n}$ = 25
$n^{2}$ = 25
$n^{}$ = 5