2. Let positive numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a^{2}(a-1)+b^{2}(b-1)+c^{2}(c-1) \\
=a(a-1)+b(b-1)+c(c-1) .
\end{array}
$$

Then $1956 a^{2}+1986 b^{2}+2016 c^{2}=$
2.5 958.

Notice that,
$$
\begin{array}{l}
x^{2}(x-1)-x(x-1)=\left(x^{2}-x\right)(x-1) \\
=x(x-1)^{2} .
\end{array}
$$

From the given information,
$$
a(a-1)^{2}+b(b-1)^{2}+c(c-1)^{2}=0 \text {. }
$$

Since $a$, $b$, and $c$ are positive numbers, it can only be that $a=b=c=1$, so,
$$
\begin{array}{l}
1956 a^{2}+1986 b^{2}+2016 c^{2} \\
=1956+1986+2016=5958 .
\end{array}
$$