Solving cubic equations

Assume that $f(x) = x^{3} + ax^{2} + bx + c = 0$ , where a, b, and c are Real and that the roots of f(x) = 0 are smooth (i.e. differentiable) functions of at least ‘c’ and are distinct. Denote the root functions by x₁, x₂ and x₃.

If x is any one of these roots, it is easy to show that ∂x/∂c=-1/f ‘(x).

Also, ∆²=(x₁-x₂)²(x₂-x₃)²(x₁-x₃)²can be shown to be independent of the roots x₁, x₂and x₃and that it is just a function of a, b, and c. $\Delta ^{2}$ can also be shown to satisfy the equation $\Delta ^{2} = -f'(x_{1} )f'(x_{2} )f'(x_{3} )$ .

Now $\Delta ^{2} (\partial x_{1}/\partial c)^{2} +f'(x_{1} )=(x_{2} -x_{3} )^{2} +(x_{1} -x_{2} )(x_{1} -x_{3} )$ .

The right-hand side of this equation simplifies to $a^{2} - 3b$ , i.e. it is independent of c. Hence the left-hand side can be differentiated with respect to c to give 0. If we do this, and use the fact that $\partial x_{1} /\partial c \ne 0$ , we get the final LDE, (which is true for any root x)

$2\Delta ^{2} \partial ^{2} x/\partial c^{2} +(\partial \Delta ^{2} /\partial c)\partial x/\partial c+f''(x)=0$ . It is straightforward to show that $\Delta ^{2} =-(27(c-ab/3+2a^{3} /27)^{2} +4(b-a^{2}/3)^{3} )$ .

If we now apply this to the reduced cubic equation f(x)=x³+ax+b=0, then it follows that

$\Delta ^{2} =-(4a^{3} +27b^{2} )$ .

The LDE above then boils down to $(4a^{3} +27b^{2} )\partial ^{2} x/\partial b^{2} +27b\partial x/\partial b-3x=0$ .

Solution of the above LDE.

Case 1 where $4a^{3} +27b^{2} <0$ , a<0 and $b\in (-\sqrt{-4a^{3} /27} ,\sqrt{-4a^{3} /27})$ .

The initial conditions for the LDE come from the reduced cubic with b=0,

i.e. $x^{3} +ax=0$ so $b=0\Rightarrow x=0, x=\sqrt{-a}, x=-\sqrt{-a}$ .

If b=0 and x=0, then $\partial x/\partial b=-1/a$ . If b=0 and $x=\pm \sqrt{-a}$ then

$\partial x/\partial b=1/2a$ . These initial conditions are enough to allow the LDE to be solved using standard methods to give …

$x_{1} =2\sqrt{-a/3} sin(1/3sin^{-1}(3b\sqrt{3/-4a^{3} } ) )$ ,

$x_{2} =2\sqrt{-a/3} sin(1/3(sin^{-1}(3b\sqrt{3/-4a^{3} } )+2\pi ))$ , and

$x_{3} =2\sqrt{-a/3} sin(1/3(sin^{-1}(3b\sqrt{3/-4a^{3} } )+4\pi ))$ .

Case 2 where $4a^{3} +27b^{2} >0$ and a>0.

If b=0, then from the reduced cubic we get one real value for x, i.e. x=0, so

$\partial x/\partial b=-1/a$ when b=0.

Again solving the LDE using standard methods gives the Cardano solution …

$x=\sqrt[3]{(-b/2+\sqrt{b^{2} /4+a^{3} /27} )}+\sqrt[3]{(-b/2-\sqrt{b^{2} /4+a^{3} /27} )}$ .

This is also a valid solution if a<=0. The complex valued solutions can also be obtained in a similar way, i.e. when b = 0 take $x = \pm i\sqrt{a}$ and $\partial x/\partial b=1/2a$ .