jmc

Let $$p(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0,$$where $a_n\ne 0.$ Then $$\begin{array}{rl}p(x^3)-p(x^3-2)& =a_n{x}^{3n}+a_{n-1}{x}^{3n-3}+\cdots -a_n(x^3-2{)}^n-a_{n-1}(x^3-2{)}^{n-1}-\cdots \\ & =a_n{x}^{3n}+a_{n-1}{x}^{3n-3}+\cdots -a_n{x}^{3n}-2n{a}_n{x}^{3n-3}-\cdots -a_{n-1}{x}^{3n-3}-\cdots \\ & =2n{a}_n{x}^{3n-3}+\cdots .\end{array}$$Thus, the degree of $p(x^3)-p(x^3-2)$ is $3n-3.$

The degree of $[p(x{)}^2]+12$ is $2n,$ so $3n-3=2n,$ which means $n=3.$

Let $p(x)=a{x}^3+b{x}^2+cx+d.$ Then $$\begin{array}{rl}p(x^3)-p(x^3-2)& =a{x}^9+b{x}^6+c{x}^3+d-(a(x^3-2{)}^3+b(x^3-2{)}^2+c(x^3-2)+d)\\ & =6a{x}^6+(-12a+4b)x^3+8a-4b+2c,\end{array}$$and $$[p(x){]}^2+12=a^2{x}^6+2ab{x}^5+(2ac+b^2)x^4+(2ad+2bc)x^3+(2bd+c^2)x^2+2cdx+d^2+12.$$Comparing coefficients, we get $$\begin{array}{rl}{a}^2& =6a,\\ 2ab& =0,\\ 2ac+b^2& =0,\\ 2ad+2bc& =-12a+4b,\\ 2bd+c^2& =0,\\ 2cd& =0,\\ d^2+12& =8a-4b+2c.\end{array}$$From the equation $a^2=6a,$ $a=0$ or $a=6.$ But since $a$ is a leading coefficient, $a$ cannot be 0, so $a=6.$

From the equation $2ab=0,$ $b=0.$

Then the equation $2ac+b^2=0$ becomes $12c=0,$ so $c=0.$

Then the equation $2ad+2bc=-12a+4b$ becomes $12d=-72,$ so $d=-6.$ Note that $(a,b,c,d)=(6,0,0,-6)$ satisfies all the equations.

Therefore, $p(x)=\boxed{6x^3-6}.$