Let's determine the sum of the fourth powers of two numbers, given that the sum of these numbers is 10 and their product is 4.
I. solution. It is not difficult to determine the two numbers in question - let's denote these by $x$ and $y$ - and then the sum of their fourth powers. According to the problem,

$$
x+y=10, \quad x y=4
$$

thus $x$ and $y$ are the two roots of the

$$
z^{2}-10 z+4=0
$$

quadratic equation, that is, $+5+\sqrt{21}$ and $+5-\sqrt{21}$ (we can consider either as $x$, the other as $y$). The sum of their fourth powers is

$$
\begin{aligned}
& (5+\sqrt{21})^{4}+(5-\sqrt{21})^{4}=\left((5+\sqrt{21})^{2}\right)^{2}+\left((5-\sqrt{21})^{2}\right)^{2}= \\
& =(46+10 \sqrt{21})^{2}+(46-10 \sqrt{21})^{2}=2\left(46^{2}+10^{2} \cdot 21\right)=8432
\end{aligned}
$$

II. solution. The problem can also be solved more simply without calculating the two numbers. The sum of the fourth powers of two numbers can generally be expressed using the sum of the two numbers - let's denote it by $p$ - and their product - let's denote it by $q$:

$$
\begin{gathered}
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=\left((x+y)^{2}-2 x y\right)^{2}-2(x y)^{2}= \\
=\left(p^{2}-2 q\right)^{2}-2 q^{2}=p^{4}-4 p^{2} q+2 q^{2}
\end{gathered}
$$

Substituting the data $p=10$ and $q=4$:

$$
x^{4}+y^{4}=8432
$$

Remarks. 1. The sum of the fourth powers of the two numbers can be written in several ways using their sum and product, for example,

$$
\begin{aligned}
(x+y)^{4} & =x^{4}+y^{4}+4 x^{3} y+4 x y^{3}+6 x^{2} y^{2}=x^{4}+y^{4}+4 x y\left(x^{2}+y^{2}\right)+ \\
& +6(x y)^{2}=x^{4}+y^{4}+4 x y\left((x+y)^{2}-2 x y\right)+6(x y)^{2}
\end{aligned}
$$

From this, expressing $x^{4}+y^{4}$ and rearranging, we again obtain the above expression.

2. The polynomial $x^{4}+y^{4}$ does not change if $x$ and $y$ are swapped, the same holds for the polynomials $x+y$ and $x y$. Generally, a polynomial in the variables $x_{1}, x_{2}, \ldots, x_{n}$ is called a symmetric polynomial if the polynomial does not change (except possibly for the order of the terms) when the variables are replaced by the same variables in any order. If $k \leq n$, the sum of all $k$-factor products that can be formed from the variables $x_{1}, x_{2}, \ldots, x_{n}$ is called the ( $k$-th degree) elementary symmetric polynomial of the variables. According to a famous theorem of algebra, every symmetric polynomial can be expressed as a polynomial in the elementary symmetric polynomials of its variables. This expression (in condensed form) is unique up to the order of the terms.

This is the expression we produced in the II. solution, and it is not coincidental that the transformation in the 1st remark also leads to the same result.