Cubic formula

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The cubic formula gives the roots of any cubic equation

${\begin{array}{l}\displaystyle {ax^{3}+bx^{2}+cx+d=0,\quad a\neq 0.}\end{array}}$

The three (distinct or not) roots are given by (Cardano published a similar formula in 1545)

$x=p+{\sqrt[{3}]{q-Q}}+{\sqrt[{3}]{q+Q}},\,$

where

$Q={\sqrt {q^{2}+(r-p^{2})^{3}}},\,$

$p=-{\frac {b}{3a}},\,$

$q=p^{3}-\left({\frac {pc+d}{2a}}\right),\,$

$r={\frac {c}{3a}}.\,$

Equivalently, the three (distinct or not) roots of

${\begin{array}{l}\displaystyle {x^{3}+{\tfrac {b}{a}}\,x^{2}+{\tfrac {c}{a}}\,x+{\tfrac {d}{a}}=0=x^{3}-3Bx^{2}+3Cx-D,\quad a\neq 0,}\end{array}}$

with 

B =  −  b 3a , C = c 3a

and 

D =  −  d a

, may be written as

${\begin{array}{l}\displaystyle {x=B+{\sqrt[{3}]{q-Q}}+{\sqrt[{3}]{q+Q}},}\end{array}}$

where

${\begin{array}{l}\displaystyle {Q={\sqrt {q^{2}+(C-B^{2})^{3}}},}\end{array}}$

${\begin{array}{l}\displaystyle {q=B^{3}-\left({\frac {3BC-D}{2}}\right).}\end{array}}$

Completing the cube

The three roots are obtained by completing the cube, i.e.

${\begin{array}{l}\displaystyle {\left(x+{\tfrac {b}{3a}}\right)^{3}-3\left({\tfrac {b}{3a}}\right)^{2}\,x-\left({\tfrac {b}{3a}}\right)^{3}+{\tfrac {c}{a}}\,x+{\tfrac {d}{a}}=0=\left(x-B\right)^{3}-3B^{2}x+B^{3}+3Cx-D,\quad a\neq 0,}\end{array}}$

or, letting 

y = x + b 3a = x  −  B

,

${\begin{array}{l}\displaystyle {y^{3}-3\left({\tfrac {b}{3a}}\right)^{2}\,\left(y-{\tfrac {b}{3a}}\right)-\left({\tfrac {b}{3a}}\right)^{3}+{\tfrac {c}{a}}\,\left(y-{\tfrac {b}{3a}}\right)+{\tfrac {d}{a}}=0=y^{3}-3B^{2}(y+B)+B^{3}+3C(y+B)-D,\quad a\neq 0,}\end{array}}$

yielding the depressed cubic equation

${\begin{array}{l}\displaystyle {y^{3}+3(C-B^{2})y-2B^{3}+3BC-D=0=y^{3}+3Ry+S,}\end{array}}$

with 

R = C  −  B 2

and 

S =  −  2B 3  +  3BC  −  D

.

Solving the depressed cubic equation with Vieta's substitution

The depressed cubic equation

$y^{3}+3Ry+S=0\,$

is solved by doing Vieta's substitution

$y=w-{\frac {R}{w}},\,$

yielding

$w^{3}+S-{\frac {R^{3}}{w^{3}}}=0.$

Multiplying by 

w 3

, it becomes a sextic equation in 

w

, which happens to be a quadratic equation in 

z = w 3

$w^{6}+Sw^{3}-R^{3}=0=z^{2}+Sz-R^{3},\,$

which is solved with the quadratic formula to get

${\begin{aligned}w^{3}&={\frac {-S\pm {\sqrt {S^{2}+4R^{3}}}}{2}}={\frac {(2B^{3}-3BC+D)\pm {\sqrt {(2B^{3}-3BC+D)^{2}+4(C-B^{2})^{3}}}}{2}}\\&=q\pm {\sqrt {q^{2}+(C-B^{2})^{3}}}=q\pm Q,\end{aligned}}\,$

with

${\begin{array}{l}\displaystyle {Q={\sqrt {q^{2}+(C-B^{2})^{3}}},}\end{array}}$

${\begin{array}{l}\displaystyle {q=B^{3}-\left({\frac {3BC-D}{2}}\right).}\end{array}}$

If 

w1  ±

, 

w2  ±

and 

w3  ±

are the three cube roots for each of the two solutions in 

w 3

, then

${y_{i}}_{\pm }={w_{i}}_{\pm }-{\frac {R}{{w_{i}}_{\pm }}}={w_{i}}_{\pm }-\left({\frac {C-B^{2}}{{w_{i}}_{\pm }}}\right),\quad 1\leq i\leq 3,\,$

and

${\begin{array}{l}\displaystyle {{x_{i}}_{\pm }=B+{y_{i}}_{\pm }=B+{w_{i}}_{\pm }-{\frac {R}{{w_{i}}_{\pm }}}=B+{w_{i}}_{\pm }-\left({\frac {C-B^{2}}{{w_{i}}_{\pm }}}\right),\quad 1\leq i\leq 3.}\end{array}}$

Thus, the three (distinct or not) roots are given by (see how the three roots of the two solutions are used in pairs, yielding three roots)

${\begin{array}{l}\displaystyle {x=B+{\sqrt[{3}]{q\pm Q}}+\left({\frac {B^{2}-C}{\sqrt[{3}]{q\pm Q}}}\right)=B+{\sqrt[{3}]{q\pm Q}}+{\sqrt[{3}]{q\mp Q}},}\end{array}}$

where 

3√  q  ∓  Q  = B 2  −  C 3√  q  ±  Q

, since 

q 2  −  Q 2 = (B 2  −  C ) 3

.
Finally, the three (distinct or not) roots of 

a x 3 + b x 2 + c x + d = 0, a   ≠   0,

may be written as

${\begin{array}{l}\displaystyle {x=B+{\sqrt[{3}]{q-Q}}+{\sqrt[{3}]{q+Q}},}\end{array}}$

where

${\begin{array}{l}\displaystyle {Q={\sqrt {q^{2}+(C-B^{2})^{3}}},}\end{array}}$

${\begin{array}{l}\displaystyle {q=B^{3}-\left({\frac {3BC-D}{2}}\right),}\end{array}}$

with 

B =  −  b 3a , C = c 3a

and 

D =  −  d a

.

Vieta's formulas for the cubic

Vieta's formulas for the cubic

$(x-x_{1})(x-x_{2})(x-x_{3})=0,\,$

gives a system of three equations in three variables (which are the three roots)

${\begin{array}{l}\displaystyle {\begin{aligned}x_{1}+x_{2}+x_{3}&=-{\frac {b}{a}}=3B,\\x_{1}\cdot x_{2}+x_{1}\cdot x_{3}+x_{2}\cdot x_{3}&=+{\frac {c}{a}}=3C,\\x_{1}\cdot x_{2}\cdot x_{3}&=-{\frac {d}{a}}=D.\end{aligned}}\end{array}}$

By dividing the third equation into the second equation, one obtains a formula for the harmonic sum of the roots

${\begin{array}{l}\displaystyle {{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}+{\frac {1}{x_{3}}}=-{\frac {c}{d}}=3{\frac {C}{D}}.}\end{array}}$

See also

• Quadratic formula
• Quartic formula or biquadratic formula

External links

• The Cubic Formula (Solve Any 3rd Degree Polynomial Equation)