Solution of a cubic equation (Part 2)

In the post Solution of a cubic equation Part 1 we found the solution of a cubic equation for a particular condition. Then, in the post Whether real or complex (Cubic Equation) we studied when the roots of a cubic equation will be real and when it will be complex. This post deals with the solution of cubic equation. The algebraic method of solving cubic equations is supposed to be due to the Italian, del Ferro (1465-1526). But it is called Cardano's method because it became known to people after the Italian, Girolamo Cardano, published it in 1545 in his 'Ars Magna'.

Omar Khayyam gave a great deal of thought to the cubic equations. Before him, Greek mathematicians obtained solutions for third degree equations by considering geometric methods that involved the intersection of conics. Although the solution is present but I am searching for a solution of cubic equation with almost a new perspective. I will continue my search but I am giving this method so that until I find my kind of solution the traditional solution is present on my blog.

Let us consider the equation (ax³ + bx² + cx + d = 0 ; a≠0)

Step 1: Express ax³ + bx² + cx + d = 0
as x³ + px² + qx + r = 0

Step 2: Shift the middle point of the curve on the axis x=−p/3.
Read post Solution of a cubic equation. A special solution and Why -b/3a? for detail.
x³ + 3(p/3)x² + 3(p²/9)x + p³/27 − 3(p²/9)x − p³/27 +qx+ r = 0
(x + p/3)³ − [3(p²/9) − q]x − p³/27 + r = 0
(x + p/3)³ − [p²/3 − q]x − p³/27 + r = 0


Step 3:  Let x + p/3 = y then x = y − p/3.
Substitute y and (y − p/3) for (x + p/3) and x in the above equation.
y³ +[q − p²/3](y − p/3) + r − p³/27 = 0
y³ +[q − p²/3]y − [qp/3 − p³/9 − r + p³/27] = 0
y³ +[q − p²/3]y + [r − qp/3 + 2p³/27] = 0
y³ +[q − p²/3]y + [2p³/27 − qp/3 + r] = 0

Step 4: Now the equation is of the form y³ + Ay + B = 0
where A = q − p²/3 and B = 2p³/27 − qp/3 + r
Let y = (s − A/3s)
then s³ − A³/(3s)³ − As + A²/3s + As − A²/3s + B = 0
s³ − A³/(3s)³ + B = 0
Multiplying throughout by s³, we get
s⁶ + Bs³ − A³/27 = 0
Let, s³ = z
z² + Bz − A³/27 = 0
z = [−B ±√(B² + 4A³/27)]/2

s₁³ = [−B +√(B² + 4A³/27)]/2
As we know that any equation has three cube roots. Let its cube root be s₁. then the roots are s₁, ωs₁ and ω²s₁.
Similarly,
s₂³ = [−B −√(B² + 4A³/27)]/2
Let its cube root be s₂. then the roots are s₂, ωs₂ and ω²s₂.
1/s₁³ = −(3s₂/A)³
⇒ 1/s₁³ = 1/[−B +√(B² + 4A³/27)]/2
= {[−B − √(B² + 4A³/27)]/2}/{[−B −√(B² + 4A³/27)][−B +√(B² + 4A³/27)]/4}
= {[−B − √(B² + 4A³/27)]/2}/{−4A³/(27×4)}
= {[−B − √(B² + 4A³/27)]/2}/{A³/27)}
= −(3s₂/A)³
s₁s₂ = −A/3
If we consider s₁ then −A/3s₁ = s₂.
The product of s₁ and s₂ is real so the possibilities of the roots for y (= s − A/3s) are
s₁ + s₂, (ωs₁ + ω²s₂) and (ω²s₁ + ωs₂).

Step 5: Shift the graph to the original position for the roots.
The three roots are
s₁ + s₂ − p/3, (ωs₁ + ω²s₂ − p/3) and (ω²s₁ + ωs₂ − p/3).

The solution of the equation
x³ + px² + qx + r = 0 is
s₁ + s₂ − p/3, (ωs₁ + ω²s₂ − p/3) and (ω²s₁ + ωs₂ − p/3).

Where ω is a cube root of 1 i.e (−1 + i√3)/2
Where s₁ is a cube root of [−B +√(B² + 4A³/27)]/2 and
s₂ is a cube root of [−B − √(B² + 4A³/27)]/2.
and
where A = q − p²/3 and B = 2p³/27 − qp/3 + r