关于三次方程的一种解法的疑问

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如图，*1：为什么可以这样换元，这样做的背景或者依据是什么
*2：如何由x1直接得出x2和x3的?
感谢各位佬，想学习研究一下这种解法

hbghlyj

forum.php?mod=viewthread&tid=9463
The linear map $z \mapsto a z+b \bar{z}$ maps $1,\omega,\bar\omega$ to the three roots of the cubic.
See Figure 5

hbghlyj

math.stackexchange.com/questions/4148260

For a quadratic $f(x)$ with roots $x_1, x_2$, he studies the quantity $p(x_1, x_2) = x_1 - x_2$. One observes that it takes two values on permutation of the roots, since $p(x_1, x_2) = -p(x_2, x_1)$. Then one can construct an additional polynomial

$$g(z) = (z-p(x_1, x_2))(z-p(x_2, x_1)) = z^2 - p(x_1, x_2)^2$$

which is symmetric in $x_1, x_2$ (since by construction we include every permutation, and thus permitting $x_1, x_2$ leads to just a shuffling of linear terms). Since $g(z)$ is symmetric in $x_1, x_2$, its coefficients are symmetric in $x_1, x_2$ and thus may be expressed in terms of the elementary symmetric polynomials in $x_1, x_2$ (i.e. the coefficients of our original $f(x)$) by the the fundamental theorem on symmetric polynomials. Furthermore, solving $g(z)=0$–an easier difference of squares then the general quadratic–gives us values for $p(x_1, x_2)$ and $p(x_2, x_1)$ in terms of these elementary symmetric polynomials (i.e. original coefficients of $f(x)$). Then we can solve this linear system to find $x_1, x_2$ in terms of our original coefficients of $f(x)$.

For a cubic $f(x)$, he follows much the same procedure. We now investigate the quantity $p(x_1, x_2, x_3) = x_1 + \omega x_2 + \omega^2 x_3$ where $\omega$ is the cube root of unity, observing it takes six values given the six permutations of $x_1, x_2, x_3$. Then we construct

$$g(z) = \prod (z-p(x_i, x_j, x_k)).$$

a 6th-degree polynomial symmetric in $x_1, x_2, x_3$, but one that happens to be quadratic in $z^3$ which we know how to solve. In fact, this is why a square root shows up in Cardano's method. Again, $g(z)$ has yielded an easier equation to solve, which we call the Lagrange resolvent, that is vital to finding solutions to $f(x)$.

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A symmetric polynomial in $r$, $s$, and $t$ will be a polynomial in the elementary symmetric functions $0$, $p$, and $q$, i.e. a polynomial in $p$ and $q$. You need to to know those polynomials explicitly to solve the quadratic equation explicitly. This involves getting your hands dirty.

You can see this worked out carefully in this paper by Svante Janson (pp. 1-6). What he calls $u$ and $v$ are, up to scalar factors, what you call $\lambda$ and $\mu$. In his notation, $u+v$ is one of the roots of your polynomial, $uv=-p/3$ (so $u^3v^3=-p^3/27$), and $u^3+v^3$ is the product of the roots, i.e. $-q$. This will give you your explicit quadratic polynomial with roots $u^3$ and $v^3$. Janson takes a slightly different approach, showing that $u^3-v^3=\sqrt{-\Delta/27}$ where the discriminant $\Delta=-4p^3-27q^2$. You can get $u^3$ and $v^3$ directly from those two equations.

To convert the notation, I believe that $v=\lambda/3$ and $u=w^2\mu/3$ if you take $r=\beta_1$,$s=\beta_2$,and $t=\beta_3$.

A good historical reference is Chapter 6 or William Dunham's "Journey through Genius," which covers the discovery of "Cardano's formula," which is what this is... without the Galois theory, because, well he did it about 250 years before Galois. It was one of the first significant results in mathematics that went beyond the legacy of the Greeks.

Solving the Cubic and Quartic, Aaron Landesman
The Fundamental Theorem on Symmetric Polynomials: History's First Whiff of Galois Theory

其妙

你还可以把四次方程的解法也写出来的