This article discusses a way to solve special cubic equations in the form of
a
x
3
+
b
x
2
+
c
x
+
d
=
0
,
where
c
=
b
2
3
a
{\displaystyle ax^{3}+bx^{2}+cx+d=0,\quad {\text{where }}c={\frac {b^{2}}{3a}}}
If the cubic equation satisfies that condition, then you can use the special cubic formula to find the value of
x
{\displaystyle x}
x
=
−
b
+
b
3
−
27
a
2
d
3
3
a
{\displaystyle x={\frac {-b+{\sqrt[{3}]{b^{3}-27a^{2}d}}}{3a}}}
start with
a
x
3
+
b
x
2
+
c
x
+
d
=
0
{\displaystyle ax^{3}+bx^{2}+cx+d=0}
1.) subtract
d
{\displaystyle d}
a
{\displaystyle a}
x
3
+
b
a
x
2
+
c
a
x
=
−
d
a
{\displaystyle x^{3}+{\frac {b}{a}}x^{2}+{\frac {c}{a}}x=-{\frac {d}{a}}}
2.) find the value of
k
{\displaystyle k}
x
3
+
3
k
x
2
+
3
k
2
x
+
k
3
=
(
x
+
k
)
3
{\displaystyle x^{3}+3kx^{2}+3k^{2}x+k^{3}=(x+k)^{3}}
There’s a problem with this that puts a limitation on the values of
b
{\displaystyle b}
c
{\displaystyle c}
b
3
a
{\displaystyle {\frac {b}{3a}}}
c
3
a
{\displaystyle {\sqrt {\frac {c}{3a}}}}
c
=
b
2
3
a
{\displaystyle \quad c={\frac {b^{2}}{3a}}}
If this condition is true, then the value of
k
{\displaystyle k}
b
3
a
{\displaystyle {\frac {b}{3a}}}
3.) add
(
b
3
a
)
3
{\displaystyle \left({\frac {b}{3a}}\right)^{3}}
k
3
{\displaystyle k^{3}}
x
3
+
b
a
x
2
+
c
a
x
+
b
3
27
a
3
=
−
d
a
+
b
3
27
a
3
{\displaystyle x^{3}+{\frac {b}{a}}x^{2}+{\frac {c}{a}}x+{\frac {b^{3}}{27a^{3}}}=-{\frac {d}{a}}+{\frac {b^{3}}{27a^{3}}}}
4.) factor the left side of the equation
(
x
+
b
3
a
)
3
=
−
d
a
+
b
3
27
a
3
{\displaystyle \left(x+{\frac {b}{3a}}\right)^{3}=-{\frac {d}{a}}+{\frac {b^{3}}{27a^{3}}}}
5.) rearrange the right side of the equation
(
x
+
b
3
a
)
3
=
b
3
−
27
a
2
d
27
a
3
{\displaystyle \left(x+{\frac {b}{3a}}\right)^{3}={\frac {b^{3}-27a^{2}d}{27a^{3}}}}
6.) take the cubic root of both sides of the equation
(
x
+
b
3
a
)
=
b
3
−
27
a
2
d
3
3
a
{\displaystyle \left(x+{\frac {b}{3a}}\right)={\frac {\sqrt[{3}]{b^{3}-27a^{2}d}}{3a}}}
7.) subtract
b
3
a
{\displaystyle {\frac {b}{3a}}}
x
=
b
3
−
27
a
2
d
3
3
a
−
b
3
a
{\displaystyle x={\frac {\sqrt[{3}]{b^{3}-27a^{2}d}}{3a}}-{\frac {b}{3a}}}
8.) simplify the equation
x
=
−
b
+
b
3
−
27
a
2
d
3
3
a
{\displaystyle x={\frac {-b+{\sqrt[{3}]{b^{3}-27a^{2}d}}}{3a}}}
As stated above, this formula can only be used in special cases where
c
{\displaystyle c}
b
{\displaystyle b}
c
=
b
2
3
a
{\displaystyle c={\frac {b^{2}}{3a}}}
or equivalently
b
=
±
3
a
c
{\displaystyle b=\pm {\sqrt {3ac}}}
If the cubic equation in question does not obey these equations, then a much longer formula must be used to find the solution. These two equations also restrict the cubic formula to cubic equations that only have one solution.
Example 1:
3
x
3
+
6
x
2
+
4
x
+
9
=
0
{\displaystyle 3x^{3}+6x^{2}+4x+9=0}
Step 1: Check if the equation obeys the limitations
c
=
4
=
b
2
3
a
=
6
2
3
⋅
3
=
4
{\displaystyle c=4\quad =\quad {\frac {b^{2}}{3a}}={\frac {6^{2}}{3\cdot 3}}=4}
Step 2: Since the equation obeys the criteria of a special cubic equation, the special cubic formula may be applied
x
=
−
6
+
6
3
−
27
⋅
3
2
⋅
9
3
3
⋅
3
=
−
6
−
1971
3
9
≈
−
2.05978
{\displaystyle x={\frac {-6+{\sqrt[{3}]{6^{3}-27\cdot 3^{2}\cdot 9}}}{3\cdot 3}}={\frac {-6-{\sqrt[{3}]{1971}}}{9}}\approx -2.05978}
Step 3: Check the answer
3
(
−
2.05978
)
3
+
6
(
−
2.05978
)
2
+
4
(
−
2.05978
)
+
9
≈
0
{\displaystyle 3(-2.05978)^{3}+6(-2.05978)^{2}+4(-2.05978)+9\approx 0}
Example 2:
3
x
3
+
21
x
2
+
2
x
+
3
=
0
{\displaystyle 3x^{3}+21x^{2}+2x+3=0}
Step 1: Check if the equation obeys the limitations
c
=
2
≠
b
2
3
a
=
21
2
3
⋅
3
=
49
{\displaystyle c=2\quad \neq \quad {\frac {b^{2}}{3a}}={\frac {21^{2}}{3\cdot 3}}=49}
This equation doesn’t obey the limitations, so it is not a special cubic equation.
Example 3:
3
x
3
−
6
x
2
+
4
x
−
5
=
0
{\displaystyle 3x^{3}-6x^{2}+4x-5=0}
Step 1: Check if the equation obeys the limitations
c
=
4
=
b
2
3
a
=
(
−
6
)
2
3
⋅
3
=
4
{\displaystyle c=4\quad =\quad {\frac {b^{2}}{3a}}={\frac {(-6)^{2}}{3\cdot 3}}=4}
Step 2: Since the equation obeys the criteria of a special cubic equation, the special cubic formula may be applied
x
=
6
+
−
6
3
−
27
⋅
3
2
⋅
−
5
3
3
⋅
3
=
6
+
999
3
9
≈
1.7774
{\displaystyle x={\frac {6+{\sqrt[{3}]{-6^{3}-27\cdot 3^{2}\cdot -5}}}{3\cdot 3}}={\frac {6+{\sqrt[{3}]{999}}}{9}}\approx 1.7774}
Step 3: Check the answer
3
(
1.7774
)
3
−
6
(
1.7774
)
2
+
4
(
1.7774
)
−
5
≈
0
{\displaystyle 3(1.7774)^{3}-6(1.7774)^{2}+4(1.7774)-5\approx 0}
![{\displaystyle x={\frac {-b+{\sqrt[{3}]{b^{3}-27a^{2}d}}}{3a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72c3e70f287b7d65c4175d4b53c2a630eb84c3ec)
![{\displaystyle \left(x+{\frac {b}{3a}}\right)={\frac {\sqrt[{3}]{b^{3}-27a^{2}d}}{3a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00baf24f1f31746eaa872cc63af580ecf44963c1)
![{\displaystyle x={\frac {\sqrt[{3}]{b^{3}-27a^{2}d}}{3a}}-{\frac {b}{3a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff9e5937ee6468191043637c358f030134e7c0c8)
![{\displaystyle x={\frac {-6+{\sqrt[{3}]{6^{3}-27\cdot 3^{2}\cdot 9}}}{3\cdot 3}}={\frac {-6-{\sqrt[{3}]{1971}}}{9}}\approx -2.05978}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e87086112a8139f3ef9fbfa591885e1ca1bd766)
![{\displaystyle x={\frac {6+{\sqrt[{3}]{-6^{3}-27\cdot 3^{2}\cdot -5}}}{3\cdot 3}}={\frac {6+{\sqrt[{3}]{999}}}{9}}\approx 1.7774}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f32d82cec325fee2454b3ee3b2f56dbb6f7f8960)
