Today, while delving into a proof elucidating the derivation of the cubic formula, I encountered a stumbling block in grasping its final segment. Although restating the entire argument anew is a daunting task, I find myself compelled to do so, given my solitary study circumstances.
A third-degree equation has the form [imath]x^3+ax^2+bx+c=0.[/imath] To solve it we consider the easiest possible case, where [imath]x^3-1=0[/imath]. Obviously [imath]1[/imath] is a root and we can factorize it as [imath]x^3-1=(x-1)(x^2+x+1)[/imath]. Solving [imath]x^2+x+1=0[/imath] we obtain the roots of unity where
[math]w=-\frac{1}{2}+i\frac{\sqrt3}{2}[/math]It's easy to show that [math]w^2= -1/2-i\sqrt3/2[/math] and we already know that [imath]w^3=1[/imath]. Now consider the equation [imath]x^3=A[/imath]. If [imath]\alpha[/imath] is a root to it, then the two other roots can be written as [imath]w\alpha[/imath] and [imath]w^2\alpha[/imath] since
[math](w^j\alpha)^3=(w^3)^j\alpha^3=1^jA=A[/math]for[imath]j=1[/imath] and [imath]2[/imath].
We return to the general equation [imath]x^3+ax^2+bx+c=0[/imath]. Define [imath]x=t+\alpha[/imath], where [imath]\alpha[/imath] is a constant and [imath]t[/imath] is a variable, via substitution into the equation we get
[math]x^3+ax^2+bx+c=t^3+(3\alpha +a)t^2+(3\alpha^2+2a\alpha+b)t+(\alpha^3+a\alpha^2+b\alpha+c)[/math]
We define [imath]\alpha[/imath] such that the coefficient of the [imath]x^2[/imath] vanishes. From [imath]3\alpha +a=0[/imath], such that [imath]\alpha=-a/3[/imath]. Substitution into the expression above gives
[math]t^3+(-\frac{a^2}{3}+b)t+\frac{2a^3}{27}-\frac{ab}{3}+c=0[/math]Define the coefficient for [imath]t[/imath] by [imath]p[/imath] and the constant term by [imath]q,[/imath] then the equation becomes [imath]t^3+px+q=0[/imath]. Hence it's enough to study third-degree equations with missing second-degree terms. To save symbols, we return to the more conventional symbol [imath]x[/imath], and consider the equation [imath]x^3+px+q=0[/imath]. To solve for the cubic roots we will use Lagrange's resolvents.
[math]L_1=\frac{1}{3}(x_1+wx_2+w^2x_3)[/math][math]L_2=\frac{1}{3}(wx_1+x_2+w^2x_3)[/math]
As the first step, we will determine a second-degree equation with roots [imath]L_1^3[/imath] and [imath]L_2^3[/imath]. The equation is given as
[math]X^2-(L_1^3+L_2^3)X+L_1^3L_2^3=0.[/math]
To solve this equation we must determine the sum and product of [imath]L_1^3[/imath] and [imath]L_2^3[/imath]. Consider the equation [imath]t^3-1=0[/imath]. As was previously shown
[math]t^3-1=(t-1)(t-w)(t-w^2)[/math]If we change the sign on [imath]t[/imath] into [imath]-t[/imath], we obtain the expression
[math]t^3+1=(t+1)(t+w)(t+w^2)[/math].
Define [imath]t=L_1/L_2[/imath] and multiply by [imath]L_2^3[/imath], this gives us
[math]L_1^3+L_2^3=(L_1+L_2)(L_1+wL_2)(L_1+w^2L_2)[/math]
Now we can start to calculate the sum and product of [imath]L_1[/imath] and [imath]L_2[/imath]. Observe that [imath]x_1+x_2+x_3=0[/imath], since the coefficient for [imath]x^2[/imath] is zero, and [imath]1+w+w^2=0[/imath].
[math]L_1+L_2=\frac{1}{3}((1+w)x_1+(1+w)x_2+2w^2x_3))=w^2x_3[/math]
In the same way, we get
[math]L_1+wL_2=wx_2, \ L_1+w^2L_2=x_1,[/math]Hence
[math]L_1^3+L_2^3=w^3x_1x_2x_3=-q[/math]
Where we again used the relationship between roots and coefficients for [imath]x^3+px+q=0[/imath] and [imath]w^3=1.[/imath] And we continue
[math]3L_13L_2=(x_1+wx_2+w^2x_3)(wx_1+x_2+w^2x_3)=w(x_1^2+x_2^2+x_3^2)+(1+w^2)(x_1x_2+x_1x_3+x_2x_3)[/math][math]=w((x_1+x_2+x_3)^2-3(x_1x_2+x_1x_3+x_2x_3))=-3wp.[/math]Therefore
[math]L_1L_2=-\frac{1}{3}wp \implies L_1^3L_2^3=-\frac{p^3}{27}.[/math]
Thus [imath]L_1^3[/imath] and [imath]L_2^3[/imath] are the roots of the resolvent equation
[math]X^2+qX-\frac{p^3}{27}=0[/math]
Completing the square gives
[math]\left(X+\frac{q}{2}\right)^2=\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2=\Delta[/math]By abusing the root sign a little, we can write the roots as
[math]L_1^3=-\frac{q}{2}+\sqrt\Delta, \ L_2^3=-\frac{q}{2}-\sqrt\Delta[/math]Then we can now define [imath]L_1[/imath] as
[math]L_1=\sqrt[3]{-\frac{q}{2}+\sqrt\Delta}[/math]
which is one of the cube roots from [imath]-q/2+\sqrt\Delta[/imath], then [imath]L_2[/imath] is determined by the relationship [imath]L_1L_2=-wp/3[/imath], which we deduced above. And we obtain
[math]x_1=L_1+w^2L_2=L_1+w^2(-\frac{wp}{3L_1})=L_1-\frac{p}{3L_1}[/math][math]=\sqrt[3]{-\frac{q}{2}+\sqrt\Delta}+\sqrt[3]{-\frac{q}{2}-\sqrt\Delta}.[/math]
By a similar argument, the other two roots become
[math]x_2=w^2\left(\sqrt[3]{-\frac{q}{2}+\sqrt\Delta}\right)+w\left(\sqrt[3]{-\frac{q}{2}-\sqrt\Delta}\right)[/math][math]x_3=w\left(\sqrt[3]{-\frac{q}{2}+\sqrt\Delta}\right)+w^2\left(\sqrt[3]{-\frac{q}{2}-\sqrt\Delta}\right)[/math]
A third-degree equation has the form [imath]x^3+ax^2+bx+c=0.[/imath] To solve it we consider the easiest possible case, where [imath]x^3-1=0[/imath]. Obviously [imath]1[/imath] is a root and we can factorize it as [imath]x^3-1=(x-1)(x^2+x+1)[/imath]. Solving [imath]x^2+x+1=0[/imath] we obtain the roots of unity where
[math]w=-\frac{1}{2}+i\frac{\sqrt3}{2}[/math]It's easy to show that [math]w^2= -1/2-i\sqrt3/2[/math] and we already know that [imath]w^3=1[/imath]. Now consider the equation [imath]x^3=A[/imath]. If [imath]\alpha[/imath] is a root to it, then the two other roots can be written as [imath]w\alpha[/imath] and [imath]w^2\alpha[/imath] since
[math](w^j\alpha)^3=(w^3)^j\alpha^3=1^jA=A[/math]for[imath]j=1[/imath] and [imath]2[/imath].
We return to the general equation [imath]x^3+ax^2+bx+c=0[/imath]. Define [imath]x=t+\alpha[/imath], where [imath]\alpha[/imath] is a constant and [imath]t[/imath] is a variable, via substitution into the equation we get
[math]x^3+ax^2+bx+c=t^3+(3\alpha +a)t^2+(3\alpha^2+2a\alpha+b)t+(\alpha^3+a\alpha^2+b\alpha+c)[/math]
We define [imath]\alpha[/imath] such that the coefficient of the [imath]x^2[/imath] vanishes. From [imath]3\alpha +a=0[/imath], such that [imath]\alpha=-a/3[/imath]. Substitution into the expression above gives
[math]t^3+(-\frac{a^2}{3}+b)t+\frac{2a^3}{27}-\frac{ab}{3}+c=0[/math]Define the coefficient for [imath]t[/imath] by [imath]p[/imath] and the constant term by [imath]q,[/imath] then the equation becomes [imath]t^3+px+q=0[/imath]. Hence it's enough to study third-degree equations with missing second-degree terms. To save symbols, we return to the more conventional symbol [imath]x[/imath], and consider the equation [imath]x^3+px+q=0[/imath]. To solve for the cubic roots we will use Lagrange's resolvents.
[math]L_1=\frac{1}{3}(x_1+wx_2+w^2x_3)[/math][math]L_2=\frac{1}{3}(wx_1+x_2+w^2x_3)[/math]
As the first step, we will determine a second-degree equation with roots [imath]L_1^3[/imath] and [imath]L_2^3[/imath]. The equation is given as
[math]X^2-(L_1^3+L_2^3)X+L_1^3L_2^3=0.[/math]
To solve this equation we must determine the sum and product of [imath]L_1^3[/imath] and [imath]L_2^3[/imath]. Consider the equation [imath]t^3-1=0[/imath]. As was previously shown
[math]t^3-1=(t-1)(t-w)(t-w^2)[/math]If we change the sign on [imath]t[/imath] into [imath]-t[/imath], we obtain the expression
[math]t^3+1=(t+1)(t+w)(t+w^2)[/math].
Define [imath]t=L_1/L_2[/imath] and multiply by [imath]L_2^3[/imath], this gives us
[math]L_1^3+L_2^3=(L_1+L_2)(L_1+wL_2)(L_1+w^2L_2)[/math]
Now we can start to calculate the sum and product of [imath]L_1[/imath] and [imath]L_2[/imath]. Observe that [imath]x_1+x_2+x_3=0[/imath], since the coefficient for [imath]x^2[/imath] is zero, and [imath]1+w+w^2=0[/imath].
[math]L_1+L_2=\frac{1}{3}((1+w)x_1+(1+w)x_2+2w^2x_3))=w^2x_3[/math]
In the same way, we get
[math]L_1+wL_2=wx_2, \ L_1+w^2L_2=x_1,[/math]Hence
[math]L_1^3+L_2^3=w^3x_1x_2x_3=-q[/math]
Where we again used the relationship between roots and coefficients for [imath]x^3+px+q=0[/imath] and [imath]w^3=1.[/imath] And we continue
[math]3L_13L_2=(x_1+wx_2+w^2x_3)(wx_1+x_2+w^2x_3)=w(x_1^2+x_2^2+x_3^2)+(1+w^2)(x_1x_2+x_1x_3+x_2x_3)[/math][math]=w((x_1+x_2+x_3)^2-3(x_1x_2+x_1x_3+x_2x_3))=-3wp.[/math]Therefore
[math]L_1L_2=-\frac{1}{3}wp \implies L_1^3L_2^3=-\frac{p^3}{27}.[/math]
Thus [imath]L_1^3[/imath] and [imath]L_2^3[/imath] are the roots of the resolvent equation
[math]X^2+qX-\frac{p^3}{27}=0[/math]
Completing the square gives
[math]\left(X+\frac{q}{2}\right)^2=\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2=\Delta[/math]By abusing the root sign a little, we can write the roots as
[math]L_1^3=-\frac{q}{2}+\sqrt\Delta, \ L_2^3=-\frac{q}{2}-\sqrt\Delta[/math]Then we can now define [imath]L_1[/imath] as
[math]L_1=\sqrt[3]{-\frac{q}{2}+\sqrt\Delta}[/math]
which is one of the cube roots from [imath]-q/2+\sqrt\Delta[/imath], then [imath]L_2[/imath] is determined by the relationship [imath]L_1L_2=-wp/3[/imath], which we deduced above. And we obtain
[math]x_1=L_1+w^2L_2=L_1+w^2(-\frac{wp}{3L_1})=L_1-\frac{p}{3L_1}[/math][math]=\sqrt[3]{-\frac{q}{2}+\sqrt\Delta}+\sqrt[3]{-\frac{q}{2}-\sqrt\Delta}.[/math]
By a similar argument, the other two roots become
[math]x_2=w^2\left(\sqrt[3]{-\frac{q}{2}+\sqrt\Delta}\right)+w\left(\sqrt[3]{-\frac{q}{2}-\sqrt\Delta}\right)[/math][math]x_3=w\left(\sqrt[3]{-\frac{q}{2}+\sqrt\Delta}\right)+w^2\left(\sqrt[3]{-\frac{q}{2}-\sqrt\Delta}\right)[/math]