Example 1 Express the symmetric polynomial in three variables $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$ as a polynomial in $\sigma_{1}, \sigma_{2}, \sigma_{3}$.
Analysis of the Lexicographic Order of Multivariate Polynomials.
Solution First, let's introduce the lexicographic order of multivariate polynomials. For two monomials $a x_{1}^{k_{1}} x_{2}^{k_{2}} \cdots x_{n}^{k_{n}}$ and $b x_{1}^{l_{1}} x_{2}^{l_{2}} \cdots x_{n}^{l_{n}}$ of an $n$-variable polynomial, if the first non-zero number in $k_{1}-l_{1}, k_{2}-l_{2}, \cdots, k_{n}-l_{n}$ is positive, then the monomial $a x_{1}^{k_{1}} x_{2}^{k_{2}} \cdots x_{n}^{k_{n}}$ is said to precede $b x_{1}^{l_{1}} x_{2}^{l_{2}} \cdots x_{n}^{l_{n}}$. Arranging the multivariate polynomial according to this precedence rule is called its lexicographic order. The first non-zero coefficient monomial after such an arrangement is called the leading term of the polynomial.

Returning to the problem. Since the leading term of $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$ is $x_{1}^{3}$, its exponent sequence is $(3,0,0)$, and $\sigma_{1}^{3-0} \sigma_{2}^{0-0} \sigma^{0}=\sigma_{1}^{3}$, we form the symmetric polynomial $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-\sigma_{1}^{3}=-3\left(x_{1}^{2} x_{2}+x_{2}^{2} x_{1}+\cdots\right)-6 x_{1} x_{2} x_{3}$. The leading term of this polynomial is $-3 x_{1}^{2} x_{2}$ with the exponent sequence $(2,1,0)$, and $\sigma_{1}^{2-1} \sigma_{2}^{1-0} \sigma_{3}^{0}=\sigma_{1} \sigma_{2}$, so we form the polynomial
$$
\begin{aligned}
& x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-\sigma_{1}^{3}+3 \sigma_{1} \sigma_{2} \\
= & -3\left(x_{1}^{2} x_{2}+x_{2}^{2} x_{1}+\cdots\right)-6 x_{1} x_{2} x_{3}+3\left(x_{1}^{2} x_{2}+x_{2}^{2} x_{1}+\cdots\right)+9 x_{1} x_{2} x_{3} . \\
= & 3 x_{1} x_{2} x_{3}=3 \sigma_{3},
\end{aligned}
$$

Thus, $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=\sigma_{1}^{3}-3 \sigma_{1} \sigma_{2}+3 \sigma_{3}$.
The above method can also be performed as follows, for comparison:
\begin{tabular}{cc} 
Exponent Sequence & Corresponding Monomial \\
$(3,0,0)$ & $\sigma_{1}^{3}$ \\
$(2,1,0)$ & $\sigma_{1} \sigma_{2}$ \\
$(1,1,1)$ & $\sigma_{3}$
\end{tabular}

According to the fundamental theorem, we should have $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=A \sigma_{1}^{3}+B \sigma_{1} \sigma_{2}+C \sigma_{3}$, where $A, B, C$ are undetermined coefficients. By taking $\left(x_{1}, x_{2}, x_{3}\right)=(1,0,0),(1,1,0),(1,1,1)$, we can determine $A=1, B=-3, C=3$.