For the equations numbered 1330 and 1403,

$$
\begin{gathered}
x^{3}-x^{2}-6 x+2=0 \quad \text { and } \\
x^{3}-13 x^{2}+40 x-4=0
\end{gathered}
$$

write down the cubic equations (I), (II), and (III) in sequence, with the coefficient of \(x^{3}\) always set to 1, having the following properties:

- The roots of equation (I) are equal to the squares of the roots of equation (2), or in other words, the fourth powers of the roots of equation (1).
- The roots of equation (II) are equal to the squares of the roots of equation (I), or in other words, the eighth powers of the roots of equation (1).
- The roots of equation (III) are similarly equal to the sixteenth powers of the roots of equation (1). (It is sufficient to determine the first four significant digits of the coefficients, which are increasing in magnitude.)

Denoting the coefficients of \(x^{2}\), \(x\), and the constant term in equation (III) by \(A\), \(B\), and \(C\) respectively, show that the following three numbers provide a good approximate value for the absolute value of the roots of equation (1):

$$
\sqrt[16]{-A}, \quad \sqrt[16]{-B / A}, \quad \sqrt[16]{-C / B}
$$

that is, each value or its negative approximately satisfies equation (1).

What is the relationship in magnitude between the absolute values of the roots obtained in sequence?
Following the solution to Exercise 1403. II, we form the equation whose roots are the squares of the roots of

$$
x^{3}+a x^{2}+b x+c=0
$$

By separating the odd and even degree terms, squaring, and rearranging,

$$
\begin{gathered}
x^{3}+b x=-\left(a x^{2}+c\right), \quad x^{6}+2 b x^{4}+b^{2} x^{2}=a^{2} x^{4}+2 a c x^{2}+c^{2}, \\
\left(x^{2}\right)^{3}-\left(a^{2}-2 b\right)\left(x^{2}\right)^{2}+\left(b^{2}-2 a c\right) x^{2}-c^{2}=0
\end{gathered}
$$

Naturally, from the value set \(a=-1, b=-6, c=2\) in (1), the coefficients of (2) are also given as follows:

$$
-\left(a^{2}-2 b\right)=-13=a_{1}, \quad b^{2}-2 a c=40=b_{1}, \quad-c^{2}=-2^{2}=c_{1}
$$

Now, the coefficient triplets of equations (I), (II), and (III) are:

$$
\begin{gathered}
-\left(a_{1}^{2}-2 b_{1}\right)=-89=a_{\mathrm{I}}, \quad b_{1}^{2}-2 a_{1} c_{1}=1496=b_{\mathrm{I}}, \quad-c_{1}^{2}=-2^{4}=c_{\mathrm{I}} ; \\
-\left(a_{\mathrm{I}}^{2}-2 b_{\mathrm{I}}\right)=-4949=a_{\mathrm{II}}, \quad-c_{\mathrm{I}}^{2}=-2^{8}=c_{\mathrm{II}} \\
b_{\mathrm{I}}^{2}-2 a_{\mathrm{I}} c_{\mathrm{I}}=2235168 \approx 2,235 \cdot 10^{6}=b_{\mathrm{II}} ; \\
-\left(a_{\mathrm{II}}^{2}-2 b_{\mathrm{II}}\right) \approx-1,982 \cdot 10^{7}=A, \quad-c_{\mathrm{II}}^{2}=-2^{16}=C \\
b_{\mathrm{II}}^{2}-2 a_{\mathrm{II}} c_{\mathrm{II}} \approx 4,996 \cdot 10^{12}=B, \quad \text{thus} \\
\left(x^{16}\right)^{3}-1,982 \cdot 10^{7}\left(x^{16}\right)^{2}+4,996 \cdot 10^{12} \cdot x^{16}-6,554 \cdot 10^{4}=0
\end{gathered}
$$

According to the statement, taking the roots using logarithms:

$$
\begin{array}{ll}
\left|x_{1}\right| \approx \sqrt[16]{1,982 \cdot 10^{7}} \approx 2,858 & \left(\lg \left|x_{1}\right| \approx 0,4561\right) \\
\left|x_{2}\right| \approx \sqrt[16]{\frac{4,996 \cdot 10^{12}}{1,982 \cdot 10^{7}}} \approx 2,175 & \left(\lg \left|x_{2}\right| \approx 0,3376\right) \\
\left|x_{3}\right| \approx \sqrt[16]{\frac{2^{16}}{4,996 \cdot 10^{12}}} \approx 0,3216 & \left(\lg \left|x_{3}\right| \approx 0,5074-1\right)
\end{array}
$$

Expanding the factored form of the cubic equation,

$$
\begin{gathered}
\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)= \\
=x^{3}-\left(x_{1}+x_{2}+x_{3}\right) x^{2}+\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right) x-x_{1} x_{2} x_{3}=0
\end{gathered}
$$

Thus, if the coefficient of \(x^{3}\) is 1, then the negative of the constant term is equal to the product of the roots, and the negative of the coefficient of \(x^{2}\) is equal to the sum of the roots. From (1),

$$
x_{1} x_{2} x_{3}=-2 \quad \text { and } \quad x_{1}+x_{2}+x_{3}=1
$$

According to the product, either one root is negative, or all three are, but the latter is impossible due to the positive sum. Therefore, we try one root as negative, specifically \(x_{2}\), as it provides a good approximation:

$$
2,858-2,175+0,322=1,005 \approx 1
$$

Indeed, the value of the left side of (1) is

$$
\begin{aligned}
& \text { at } x_{1} \approx 2,858 \text { approximately }+0,008 \\
& \text { at } x_{2} \approx-2,175 \text { approximately }+0,030 \\
& \text { at } x_{3} \approx+0,322 \text { approximately }-0,002
\end{aligned}
$$

thus a root is expected near each of these approximations.

The formulas in (3) gave the absolute values of the roots in decreasing order.

Jenö Deák (Budapest, Kölcsey F. g. IV. o. t.)

Remark. In the above, we saw a very simple example of the approximate determination of the (real) roots of an algebraic equation with real coefficients using the so-called Graeffe-Lobachevsky method. A complete description of this method would lead us far, but we will mention the following for some information. 1. The calculation was made possible by the fact that the roots are different in absolute value (this also excludes the existence of complex roots). 2. Therefore, by squaring the roots successively, the ratios of the roots of equations I-III became increasingly larger. 3. It would have been unnecessary to go further to the 32nd powers, to equation (IV), because in the calculation of the coefficients, \(A^{2}\) is negligible compared to \(2 B\), and \(B^{2}\) is negligible compared to \(2 A C\), so from equation (IV), with 32nd roots, we would have obtained the same approximate values.