calculus - Limit with unknown parameter

What possible values can take $a,b\in\mathbb{R}$ such that
$$\lim\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}+1}-\log _{5}\left( 3^{n}+4^{n}+5^{n}\right) -\sqrt[4]{n^{4}+1}\right) =1~?$$

Denote $\left( a_{n}\right)$ the expresion inside limit. My idea is to
manipulate the expresions one-by-one inside the limit and I get that

$$\begin{align*} \lim\limits_{n\rightarrow\infty}\left( a_{n}-1\right) & =\lim \limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}+1}-1-\log_{5}% 5^{n}\left( \left( \dfrac{3}{5}\right) ^{n}+\left( \dfrac{4}{5}\right) ^{n}+1\right) -\sqrt[4]{n^{4}+1}\right) \\ & =\lim\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}+1}% -1-n+\log_{5}\left( \left( \dfrac{3}{5}\right) ^{n}+\left( \dfrac{4}% {5}\right) ^{n}+1\right) -\sqrt[4]{n^{4}+1}\right) \\ & =\lim\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}% +1}-1-n-\sqrt[4]{n^{4}+1}\right) \\ & =\lim\limits_{n\rightarrow\infty}\left( \sqrt[3]{an^{3}+bn^{2}% +1}-1-2n-\left( \sqrt[4]{n^{4}+1}-n\right) \right) \\ & =\lim\limits_{n\rightarrow\infty}\left( \dfrac{an^{3}+bn^{2}+1-1} {\sqrt[3]{an^{3}+bn^{2}+1}^{2}+\sqrt[3]{an^{3}+bn^{2}+1}+1}-2n-\lim \limits_{n\rightarrow\infty}\dfrac{1}{n^{\alpha}+...}\right) \\ & =\lim\limits_{n\rightarrow\infty}\left( \dfrac{an^{3}+bn^{2}} {\sqrt[3]{an^{3}+bn^{2}+1}^{2}+\sqrt[3]{an^{3}+bn^{2}+1}+1}-2n\right) . \end{align*}$$
where I have used $\log_{5}\left( \left( \dfrac{3}{5}\right) ^{n}+\left( \dfrac{4}{5}\right) ^{n}+1\right) \rightarrow\log_{5}1=0.$

The limit $\lim\limits_{n\rightarrow\infty}\left( a_{n}-1\right)$ should be
equal with $0$, but I do not think that my last relation would imply this
thing. Am I wrong something?