Hint: Find substitution

Similar solution with Anoop Babu 's

$\begin{aligned} \sqrt[3]{24+x}+\sqrt{12-x} & \le 6 & \small \color{#3D99F6} \text{Let } u=\sqrt{12-x}, \ u^2 = 12 - x, \ x = 12 - u^2 \\ \sqrt[3]{24+12-u^2}+u & \le 6 \\ \sqrt[3]{36-u^2}+u & \le 6 \\ \sqrt[3]{(6-u)(6+u)}+u & \le 6 \\ \sqrt[3]{(6-u)(6+u)} & \le 6 - u & \small \color{#3D99F6} \text{Cubing both sides} \\ (6-u)(6+u) & \le (6 - u)^3 \\ (6 - u)^3 - (6-u)(6+u) & \ge 0 \\ (6-u)[(6 - u)^2 - 6-u] & \ge 0 \\ (6-u)[u^2-12u+36 - 6-u] & \ge 0 \\ (6-u)[u^2-13u+30] & \ge 0 \\ {\color{#3D99F6}(6-u)}(u-3)(u-10) & \ {\color{#3D99F6}\ge} \ 0 & \small \color{#3D99F6} \text{Changing sign and hence inequality} \\ (u-3){\color{#D61F06}(u-6)}(u-10) & \ {\color{#D61F06}\le} \ 0 \end{aligned}$

$\implies \begin{cases} u \le 3 & \implies \sqrt{12-x} \le 3 &\implies 3 \le x \le 12 & \small \implies 10 \text{ integral solutions} \\ 6 \le u \le 10 & \implies 6 \le \sqrt{12-x} \le 10 &\implies -24 \le x \le -88 & \small \implies 65 \text{ integral solutions} \end{cases}$

Therefore, a total of $\boxed{75}$ integral solutions.

Let u = ∛(24+x) , then -x = 24-u^3. The inequality becomes √(36-u^3 ) ≤ 6 – u. Square both sides, then we get 0 ≤ u^3 + u^2 – 12u = u(u+4)(u – 3), then -4≤u≤0, or u≥3, are the solutions.Transforming back to x variable, we have -64≤24+x≤0, or 24+x≥27. finally we have the solutions -88≤ x≤ - 24, or 3≤x≤12. So the number of integers solutions is 65+10 = 75 .