Solve the following system of equations:

$$
\begin{aligned}
a^{3} x+a^{2} y+a z+u & =1 \\
b^{3} x+b^{2} y+b z+u & =0 \\
c^{3} x+c^{2} y+c z+u & =0 \\
d^{3} x+d^{2} y+d z+u & =0
\end{aligned}
$$
First, we calculate a few determinants that we will need. The determinant of the system of equations is:

$$
D=\left|\begin{array}{llll}
a^{3} & a^{2} & a & 1 \\
b^{3} & b^{2} & b & 1 \\
c^{3} & c^{2} & c & 1 \\
d^{3} & d^{2} & d & 1
\end{array}\right|
$$

Subtract the elements of the last row from the corresponding elements of the first three rows, then:

![](https://cdn.mathpix.com/cropped/2024_05_02_109b1469a5295e62460dg-1.jpg?height=195&width=1193&top_left_y=474&top_left_x=455)

$$
\begin{aligned}
& =(a-d)(b-d)(c-d)\left|\begin{array}{lll}
a^{2} & a & 1 \\
b^{2} & b & 1 \\
c^{2} & c & 1
\end{array}\right|=(a-d)(b-d)(c-d)\left|\begin{array}{ccc}
a^{2}-c^{2} & a-c & 0 \\
b^{2}-c^{2} & b-c & 0 \\
c^{2} & c & 1
\end{array}\right|= \\
& =(a-c)(a-d)(b-c)(b-d)(c-d)\left|\begin{array}{ll}
a+c & 1 \\
b+c & 1
\end{array}\right|=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)
\end{aligned}
$$

Furthermore:

$$
\begin{aligned}
& D_{1}=\left|\begin{array}{llll}
1 & a^{2} & a & 1 \\
0 & b^{2} & b & 1 \\
0 & c^{2} & c & 1 \\
0 & d^{2} & d & 1
\end{array}\right|=\left|\begin{array}{lll}
b^{2} & b & 1 \\
c^{2} & c & 1 \\
d^{2} & d & 1
\end{array}\right|=(b-c)(b-d)(c-d)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_109b1469a5295e62460dg-1.jpg?height=196&width=718&top_left_y=1165&top_left_x=690)

$$
\begin{aligned}
& =-\left|\begin{array}{ccc}
b^{3}-d^{3} & b-d & 0 \\
c^{3}-d^{3} & c-d & 0 \\
d^{3} & d & 1
\end{array}\right|=-(b-d)(c-d)\left|\begin{array}{cc}
b^{2}+b d+d^{2} & 1 \\
c^{2}+c d+d^{2} & 1
\end{array}\right|= \\
& =-(b-d)(c-d)\left|\begin{array}{cc}
b^{2}-c^{2}+d(b-c) & 0 \\
c^{2}+c d+d^{2} & 1
\end{array}\right|=-(b-c)(b-d)(c-d)(b+c+d)
\end{aligned}
$$

Similarly:

$$
D_{3}=\left|\begin{array}{llll}
a^{3} & a^{2} & 1 & 1 \\
b^{3} & b^{2} & 0 & 1 \\
c^{3} & c^{2} & 0 & 1 \\
d^{3} & d^{2} & 0 & 1
\end{array}\right|=(b-c)(b-d)(c-d)(b c+b d+c d)
$$

and

$$
D_{4}=\left|\begin{array}{llll}
a^{3} & a^{2} & a & 1 \\
b^{3} & b^{2} & b & 0 \\
c^{3} & c^{2} & c & 0 \\
d^{3} & d^{2} & d & 0
\end{array}\right|=-(b-c)(b-d)(c-d) b c d
$$

Thus:

$$
\begin{aligned}
& x=\frac{D_{1}}{D}=\frac{1}{(a-b)(a-c)(a-d)} \\
& y=\frac{D_{2}}{D}=\frac{b+c+d}{(a-b)(a-c)(a-d)} \\
& z=\frac{D_{3}}{D}=\frac{b c+b d+c d}{(a-b)(a-c)(a-d)} \\
& u=\frac{D_{4}}{D}=-\frac{b c d}{(a-b)(a-c)(a-d)}
\end{aligned}
$$

(Imre Kürti, Eger.)

The problem was also solved by: Ádámffy E., Bartók I., Deutsch E., Deutsch I., Haar A., Hirschfeld Gy., Jánosy Gy., Kertész G., Pichler S., Pivnyik I., Riesz K., Rosenberg J., Schlesinger O., Schwemmer I., Schöffer I., Szücs A., Tandlich E.