$ 2(x+2)^3(x-3)\leq (x^2-4)(x+2)^2 $

Three students solved the inequality: $$ 2(x+2)^3(x-3)\leq (x^2-4)(x+2)^2 $$

Luke noticed that the expression $(x+2)^2$ is on both sides, so he decided to divide both sides of the inequality by $(x+2)^2$. He realized that division is only permissible by a non-zero expression. Thus he set the condition that $x\neq -2$. This led to simplified inequality: $$ 2(x+2)(x-3)\leq (x^2-4) $$ He further reduced the inequality: $$\begin{gather} 2(x^2-x-6)-x^2+4\leq 0 \cr x^2-2x-8\leq 0 \cr x_{1,2}=\frac{-(-2)\pm \sqrt{(-2)^2-4\cdot 1\cdot (-8)}}{2\cdot 1} \cr x_1=4,~x_2=-2 \end{gather} $$

He knew that the graph of the quadratic function $f(x)=x^2-2x-8$ is a parabola that opens upwards (due to the positive coefficient of the quadratic term) with $x$-intercepts at $-2$ and $4$. From this and the condition $x\neq -2$, he concluded that the inequality holds true for all real numbers within the interval: $$ x \in (-2;4 ] $$

Adam solved the inequality as follows: $$ \begin{gather} 2(x+2)^3(x-3)\leq (x^2-4)(x+2)^2 \cr 2(x+2)^3(x-3)\leq (x-2)(x+2)(x+2)^2 \end{gather} $$ He divided both sides of the inequality by $(x+2)^3$ and also set the condition that $x\neq -2$: $$ \begin{gather} 2(x-3)\leq x-2 \cr x\leq 4 \end{gather} $$ Adam claims that the solution set of the inequality is the set of all real numbers in the interval: $$ x \in (-\infty ;-2) \cup (-2;4 ] $$

Eva approached the inequality differently: $$ \begin{gather} 2(x+2)^3(x-3)\leq (x^2-4)(x+2)^2 \cr 2(x+2)^3(x-3)\leq (x-2)(x+2)(x+2)^2 \cr (x+2)^3(2(x-3)-(x-2))\leq 0 \cr (x+2)^3(x-4)\leq 0 \end{gather} $$ She noticed that the expression on the left side becomes zero when $x=-2$ and $x=4$ and that the expression is negative when $x>-2$ and $x<4$. She deduced that the inequality is satisfied if and only if: $$ x\in [ -2;4 ] $$ Which student proceeded correctly in solving the inequality?