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Cubic Equation

 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.

 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.

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



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