463. Solve the system of equations:

$$
\begin{aligned}
& z + a v + a^{2} x + a^{3} = 0 \\
& z + b y + b^{2} x + b^{3} = 0 \\
& z + c y + c^{2} x + c^{3} = 0
\end{aligned}
$$

## Bourdon Problems.
463. Binet's Solution. These equations express by their composition that $\quad U^{3}+x U^{2}+y U+z$ becomes zero three times when substituting the quantities $a, b$, and $c$ for $U$, hence it is divisible by the product $(U-a)(U-b)(U-c)$, with the quotient being one, because the first term of the divisor is $U^{3}$. We obtain the identity:

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-199.jpg?height=483&width=483&top_left_y=1186&top_left_x=1232)

Fig. 107.

$$
U^{3}+x U^{2}+y U+z=(U-a)(U-b)(U-c)
$$

or

$$
\begin{gathered}
U^{3}+x U^{2}+y U+z=U^{3}-(a+b+c) U^{2}+ \\
+(a b+a c+b c) U-a b c
\end{gathered}
$$

from which

$$
\begin{gathered}
x=-(a+b+c) \\
y=a b+a c+b c \\
z=-a b c
\end{gathered}
$$

Bourdon, a French pedagogue and mathematician, author of widely used courses in elementary mathematics. His "Éléments d'Algèbre" went through more than 20 editions, and "Éléments d'Arithmétique" up to 40 editions.