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Cubic Formula — Definition, Formula & Examples

The Cubic Formula is a closed-form expression that gives the exact roots of any cubic equation ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0, analogous to how the quadratic formula solves degree-2 equations. It was first published in the 16th century and is often called Cardano's formula after the mathematician who popularized it.

Given the depressed cubic t3+pt+q=0t^3 + pt + q = 0 (obtained from the general cubic by the substitution t=x+b3at = x + \tfrac{b}{3a}), the cubic formula states that one root is t=−q2+q24+p3273+−q2−q24+p3273t = \sqrt[3]{-\tfrac{q}{2} + \sqrt{\tfrac{q^2}{4} + \tfrac{p^3}{27}}} + \sqrt[3]{-\tfrac{q}{2} - \sqrt{\tfrac{q^2}{4} + \tfrac{p^3}{27}}}. The remaining two roots are obtained by multiplying the cube-root terms by the primitive cube roots of unity ω=e2πi/3\omega = e^{2\pi i/3} and ω2\omega^2.

Key Formula

t=−q2+q24+p3273  +  −q2−q24+p3273t = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} \;+\; \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}
Where:
  • tt = Root of the depressed cubic $t^3 + pt + q = 0$
  • pp = Coefficient of $t$ in the depressed cubic, where $p = \frac{c}{a} - \frac{b^2}{3a^2}$
  • qq = Constant term of the depressed cubic, where $q = \frac{2b^3}{27a^3} - \frac{bc}{3a^2} + \frac{d}{a}$

How It Works

To apply the cubic formula, you first reduce the general cubic ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0 to a depressed cubic t3+pt+q=0t^3 + pt + q = 0 by substituting x=t−b3ax = t - \frac{b}{3a}. This eliminates the x2x^2 term. You then compute the discriminant-like quantity Δ0=q24+p327\Delta_0 = \frac{q^2}{4} + \frac{p^3}{27}. If Δ0>0\Delta_0 > 0, there is one real root and two complex conjugate roots; if Δ0=0\Delta_0 = 0, all roots are real with at least a repeated root; if Δ0<0\Delta_0 < 0, all three roots are real and distinct (the "casus irreducibilis"), but the formula produces expressions involving complex cube roots even though the final answers are real. After finding tt, you recover x=t−b3ax = t - \frac{b}{3a}.

Worked Example

Problem: Solve x3−6x−9=0x^3 - 6x - 9 = 0 using the cubic formula.
Step 1: Identify as depressed cubic: The equation has no x2x^2 term, so it is already in the form t3+pt+q=0t^3 + pt + q = 0 with t=xt = x, p=−6p = -6, and q=−9q = -9. No substitution is needed.
x3+(−6)x+(−9)=0x^3 + (-6)x + (-9) = 0
Step 2: Compute the inner discriminant: Calculate Δ0=q24+p327\Delta_0 = \frac{q^2}{4} + \frac{p^3}{27}.
Δ0=(−9)24+(−6)327=814+−21627=814−8=494\Delta_0 = \frac{(-9)^2}{4} + \frac{(-6)^3}{27} = \frac{81}{4} + \frac{-216}{27} = \frac{81}{4} - 8 = \frac{49}{4}
Step 3: Apply the formula: Since Δ0>0\Delta_0 > 0, the square root is real: 49/4=7/2\sqrt{49/4} = 7/2.
x=92+723+92−723=83+13=2+1=3x = \sqrt[3]{\frac{9}{2} + \frac{7}{2}} + \sqrt[3]{\frac{9}{2} - \frac{7}{2}} = \sqrt[3]{8} + \sqrt[3]{1} = 2 + 1 = 3
Step 4: Verify and find remaining roots: Substituting x=3x = 3: 27−18−9=027 - 18 - 9 = 0. Factoring out (x−3)(x - 3) gives x2+3x+3=0x^2 + 3x + 3 = 0, whose roots are x=−3±i32x = \frac{-3 \pm i\sqrt{3}}{2}.
x3−6x−9=(x−3)(x2+3x+3)x^3 - 6x - 9 = (x - 3)(x^2 + 3x + 3)
Answer: The real root is x=3x = 3. The two complex roots are x=−3+i32x = \frac{-3 + i\sqrt{3}}{2} and x=−3−i32x = \frac{-3 - i\sqrt{3}}{2}.

Why It Matters

The cubic formula is a key topic in college-level abstract algebra and history of mathematics courses. Its discovery in the 16th century by del Ferro, Tartaglia, and Cardano led directly to the acceptance of complex numbers, since the formula naturally produces expressions involving −1\sqrt{-1}. Engineers and scientists who need exact symbolic roots of cubic polynomials—for instance when analyzing characteristic equations of 3×33 \times 3 matrices—rely on this result.

Common Mistakes

Mistake: Forgetting to reduce the general cubic to a depressed cubic before applying the formula.
Correction: Always substitute x=t−b3ax = t - \frac{b}{3a} first to eliminate the x2x^2 term. The formula only applies directly to the depressed form t3+pt+q=0t^3 + pt + q = 0.
Mistake: Assuming a negative value under the square root means no real roots exist.
Correction: When Δ0<0\Delta_0 < 0, all three roots are actually real. This is the casus irreducibilis. You must work with complex cube roots and the imaginary parts will cancel to give real answers.

Related Terms