us consider the following vector subspace of the algebra $\mathfrak{A}$: $$\begin{aligned}
\mathfrak{A_1}&=\overline{\text{Span}}\{\mathbb{I}, \mathbb{N}_1, \mathbb{N}_2\};
\\
\mathfrak{A_2}&=\overline{\text{Span}}\{\aspm, \arpm\};
\\
\mathfrak{A}_3^L&=\overline{\text{Span}}\{\widehat{P}^{k,\,[l]}_{m,\,[n]}\},\qquad
\mathfrak{A}_3^R=\overline{\text{Span}}\{\widehat{P}^{[k],\,l}_{[m],\,n}\},\quad k,l,m,n\geq 0;
\\
\mathfrak{A}_3&=\mathfrak{A}_3^L\boldsymbol{\dot{+}}\mathfrak{A}_3^R,
\\
\mathfrak{A}_4&=\overline{\text{Span}}\{P^{k,\,l}_{m,\,n}\},
\quad k,l,m,n\geq 0;\end{aligned}$$ $$\mathfrak{B}_1^L=\mathfrak{A_2}\boldsymbol{\dot{+}}\mathfrak{A}_3^L,\qquad
\mathfrak{B}_1^R=\mathfrak{A_2}\boldsymbol{\dot{+}}\mathfrak{A}_3^R.$$ $$\mathfrak{B}_1=\mathfrak{B}_1^L\boldsymbol{\dot{+}}\mathfrak{B}_1^R,$$ $$\mathfrak{B}_2^L=
\mathfrak{A_2}\boldsymbol{\dot{+}}\mathfrak{A}_3^L\boldsymbol{\dot{+}}\mathfrak{A}_4,\qquad
\mathfrak{B}_2^R=
\mathfrak{A_2}\boldsymbol{\dot{+}}\mathfrak{A}_3^R\boldsymbol{\dot{+}}\mathfrak{A}_4,$$ $$\mathfrak{B}_2=\mathfrak{B}_2^L\boldsymbol{\dot{+}}\mathfrak{B}_2^R,$$

$$\mathfrak{B}_3^L=\mathfrak{A}_3^L\boldsymbol{\dot{+}}\mathfrak{A}_4,\qquad
\mathfrak{B}_3^R=\mathfrak{A}_3^R\boldsymbol{\dot{+}}\mathfrak{A}_4,$$

$$\mathfrak{B}_3=\mathfrak{B}_3^L\boldsymbol{\dot{+}}\mathfrak{B}_3^R,$$

Of the commutation relation -, it follows that

-   $\mathfrak{A}_1$ — Abelian subalgebra of infinite-dimensional algebra $\mathfrak{A}$;

-   $\mathfrak{B}_2^L$ , $\mathfrak{B}_2^R$ , $\mathfrak{B}_2^L\bigoplus\mathfrak{B}_2^R$ — two-sided ideals in $\mathfrak{A}$, where $\bigoplus$ — direct sum of ideals;

-   $\mathfrak{B}_3^L$ — two-sided ideals in $\mathfrak{A}$ and therefore two-sided ideals in $\mathfrak{B}_2$ and in $\mathfrak{B}_2^L$;

-   $\mathfrak{B}_3^R$ — two-sided ideals in $\mathfrak{A}$ and therefore two-sided ideals in $\mathfrak{B}_2$ and in$\mathfrak{B}_2^R$;

-   $\mathfrak{A}_4$ — two-sided ideals in $\mathfrak{A}$ and therefore two-sided ideals in $\mathfrak{B}_2$;

Then $\mathfrak{A}_4$ is two-sided ideals in $\mathfrak{B}_2^L$ and in $\mathfrak{B}_2^R$, as well in $\mathfrak{B}_3$, $\mathfrak{B}_3^L$ and in $\mathfrak{B}_3^R$ too. We construct derived series for the algebra $\mathfrak{A}$ $$\mathfrak{A}^{(0)}=\mathfrak{A},\quad
\mathfrak{A}^{(1)}=[\mathfrak{A}^{(0)},\mathfrak{A}^{(0)}],\quad
\mathfrak{A}^{(2)}=[\mathfrak{A}^{(1)},\mathfrak{A}^{(1)}],\quad \ldots .$$ It can be checked that $$\label{13-7.28}
\mathfrak{A}^{(1)}=\mathfrak{B}_2,\qquad
\mathfrak{A}^{(n)}=\mathfrak{A}^{(2)}=\mathfrak{B}_3,\quad \forall n\geq2.$$ From it follows that $\mathfrak{A}$ is not solvable and therefore also not nilpotent algebra.

Denote by $\mathbb{N}$ a radical (i.e. maximal solvable ideal) of algebra $\mathfrak{A}$. It is known that $\mathfrak{Z}(\mathfrak{A})\subseteq \mathbb{N}$. Then, taking into account , we have 