1.6.5 $\star \star$ Find a function $p(x)$ such that $p^{(8)}(x)=x^{2}+2 x$.


Let $f(x)=x^{2}+2 x, \varphi(x)=x+1, g(x)=x^{2}$, then $\varphi^{-1}(x)=x-1$. Thus, $f(x)=\varphi^{-1}(g(\varphi(x)))$. Let $h(x)=x^{\frac{5}{2}}$, then $h^{(8)}(x)=x^{2}=g(x)$. Therefore, we take
$$
p(x)=\varphi^{-1}(h(\varphi(x)))=(x+1)^{\sqrt{2}}-1,
$$

then $p^{(8)}(x)=\varphi^{-1}\left(h^{(8)}(\varphi(x))\right)=\varphi^{-1}(g(\varphi(x)))=f(x)=x^{2}+2 x$.
Hence, $p(x)=(x+1)^{\sqrt{2}}-1$ is a function that satisfies the condition.