Cardano's Method to Solve Cubic Equations

64x³–48x²+12x–1

Cubic in normal form: x³–0.75x²+0.1875x–0.015625
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=0 q=0
So the new equation is:
t³+0t +0 [2]
As p and q are zero, then all the roots are real and equal to –a/3 (0.25)
In summary, the roots for
Cubic: 64x³–48x²+12x–1 are:
x1=0.25 (Remainder=0)
x2=0.25 (Remainder=0)
x3=0.25 (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

x³+6x²+11x+6

Cubic in normal form: x³+6x²+11x+6
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=–1 q=0
So the new equation is:
t³–1t +0 [2]
x1=–a/3, x2=–a/3± √(–p)
As p<0, then all the roots are real
In summary, the roots for
x³+6x²+11x+6 are:
x1=–2 (Remainder=0)
x2=–1 (Remainder=0)
x3=–3 (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

x³+3x²+3x–2

cubic in normal form: x³+3x²+3x–2
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=0 q=–3
So the new equation is:
t³+0t –3 [2]
As p=0, and q=–3
t= ∛(–q)=∛(3)
so x=t–a/3=∛(–q)–a/3=∛(3)–1
In summary, the roots for
Cubic: x³+3x²+3x–2 are:
x1=0.4422495703 (Remainder=0)
x2=–1.7211247852+1.2490247665i (Remainder=0)
x3=–1.7211247852–1.2490247665i (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places

1000x³–1254x²–496x+191

Cubic in normal form: x³–1.254x²–0.496x+0.191
x³+ax²+bx+c [1]
Substitute x=t–a/3, to eliminate the x² term
New equation t³+pt+q=0
Where p=b–a²/3, and q=c+ (2a³–9ab)/27
p=–1.020172 q=–0.16239726400000004
So the new equation is:
t³–1.020172t –0.16239726400000004 [2]

The discriminant of the cubic, D=(p/3)³+(q/2)²
Before rounding, D=–0.03273066871437038
D=–0.03273066871437

D> 0→ 2 complex roots
D=0 → real roots, with two the same
D<0 → Three real roots

Because D<0, all the roots are real and distinct

Our reduced equation is:
t³–1.020172t –0.16239726400000004 [2, repeated]
Or in algebra, t³+pt+q=0
There is a relationship:
(u–v)³+3uv(u–v)–(u³–v³)=0, which indicates how we might solve the equation
If we set t=u–v, p=3uv, or v=p/3u and q=–(u³–v³), that is:
v=–1.020172/(3u) [4]
u³–(–1.020172/(3*u)–0.16239726400000004=0 [5]
So u³– p³/(27u³)+q= 0
Hence, We need to find numbers, u and v, such that:
u³–v³=–q, and uv=p/3, giving t=v–u
u³–v³=0.16239726400000004, [6] and
uv=–1.020172/3 [7]
Using v=–1.020172/3u, we can eliminate u
u³– p³/27u³ =–q
The resulting equation is:
27u⁶ + 27qu³ –p³=0
Normalised, this is
u⁶ + qu³ –p³/27=0
=u⁶ +0.16239726400000004*u³ +1.020172³/27=0
This is a quadratic equation in u³, so we know how to solve it
u³=–q/2 ±√(q²/4 +p³/27) (5)
The square root may be negative, making the u's complex numbers, although the resulting roots might not be complex.
D=0.02637287135468571/4–1.0617449369321288/27=
–0.03273066871437038
After rounding, D=–0.03273066871437
The square root is a complex number, because D<0
u=∛(0.08119863200000002 ±√(0.03273066871437) i)
We need to find the cube roots of u³= 0.081198632+0.1809161925i

Convert to trigonometric form
D=(q/2)²+(p/3)³
So u³=–q/2+√(–D) i
As u³–v³=–q,
v³=q/2+√(–D) i
r is therefore the same for both u and v [A]
r²=(–q/2)²–D
=(–q/2)²–(q/2)²+(p/3)³)
=–(p/3)³)
r=√(–(p/3)³ )
Use cos to find φ, as this makes finding the angles easier (with a calculator or computer)
r=√( (1.020172/3)³) (as it is a phasor, it can have positive values only.)
r=0.1983025127249824 and r^1/3=0.5831443503398909
cos (φ)=–q/2r
cos (φ)=0.16239726400000004/(2*0.1983025127249824)
For v, cos(φ_v)=–0.16239726400000004/(2*0.1983025127249824)
So, cos(φ_v)= – cos(φ) [B]
Similarly, sin(φ_v)= sin(φ) [C]
acos(φ)= –q/(2*r)
φ=1.148924921167228
x1=u–v–a/3
u=r^1/3*(cos(φ/3) + i sin(φ/3)
v=r^1/3*(–cos(φ/3) + i sin(φ/3) (from equations B and C above)
So, x1=2*r*cos(φ)–a/3 (because the sines cancel and the cosines are doubled
x1=2*0.5831443503398909*cos(1.148924921167228)+1.254/3
We find the other x's by adding 2π to φ, and dividing the result by 3
φ₂=φ+2*π
x2=2*0.5831443503398909*cos(2.4773700761156046)––1.254/3
φ₂=φ+4*π
x3=2*0.5831443503398909*cos(4.571765178508801)––1.254/3

In summary, the roots for
Cubic: 1000x³–1254x²–496x+191 are:
x1=1.4997993055 (Remainder=0)
x2=–0.5003313644 (Remainder=0)
x3=0.254532059 (Remainder=0)
The remainder is the result of substituting the value in the equation, rounded to 10 decimal places