In short

A cubic equation has the form ax^3 + bx^2 + cx + d = 0 with a \neq 0. It always has exactly three roots (counting multiplicities and complex roots). The roots are tied to the coefficients by Vieta's formulas — the sum of roots equals -b/a, the sum of pairwise products equals c/a, and the product of all three equals -d/a. Special cubics can be solved by factoring or by spotting patterns; the general case has a closed-form solution known as Cardano's formula, which is far wilder than the quadratic formula.

Take the equation x^3 - 6x^2 + 11x - 6 = 0. Try x = 1: you get 1 - 6 + 11 - 6 = 0. Try x = 2: 8 - 24 + 22 - 6 = 0. Try x = 3: 27 - 54 + 33 - 6 = 0. All three work. A cubic equation can have three solutions — one more than a quadratic, and the jump from two to three is not just a change in quantity. It changes the character of the problem.

A quadratic equation has a single, elegant formula that produces both roots. You might expect the same pattern to continue: a clean formula for cubics, a slightly messier one for quartics, and so on up the degrees. The reality is stranger. A formula for cubics exists, but it is genuinely surprising — it requires cube roots of complex numbers even when all three roots are real. That tension between the simple-looking equation and the unexpectedly deep answer makes cubic equations one of the most interesting topics in classical algebra.

The shape of a cubic

A cubic equation is any equation you can write as

ax^3 + bx^2 + cx + d = 0

where a, b, c, d are real numbers and a \neq 0. The left side is a degree-3 polynomial, and its graph is a smooth curve with at most one hump and one dip — an S-shaped wiggle that, unlike a parabola, extends to both +\infty and -\infty.

Three types of cubic curves showing one real root, three distinct real roots, and a repeated rootThree cubic curves side by side. The first crosses the horizontal axis at one point. The second crosses at three distinct points, showing the typical S-shape. The third touches the axis at one point and crosses at another, illustrating a repeated root. one real root + two complex roots three distinct real roots repeated root
Three cubic curves. The first has one real root (and two complex ones). The second has three distinct real roots — the classic S-shape crossing the axis three times. The third touches the axis at one point (a repeated root) and crosses at another. Every cubic has at least one real root, because the curve must cross the axis at least once on its journey from $-\infty$ to $+\infty$.

Because a cubic goes from -\infty to +\infty (or +\infty to -\infty if a < 0), it must cross the horizontal axis at least once. So every cubic equation with real coefficients has at least one real root. It might have one, two (when one is repeated), or three. The Fundamental Theorem of Algebra guarantees exactly three roots in total, counting multiplicities and complex roots.

Roots and coefficients: Vieta's formulas for cubics

When you studied quadratics, you saw that the two roots \alpha, \beta of ax^2 + bx + c = 0 satisfy \alpha + \beta = -b/a and \alpha\beta = c/a. The same pattern extends to cubics, with one more relationship.

If \alpha, \beta, \gamma are the three roots of ax^3 + bx^2 + cx + d = 0, then:

Vieta's formulas for a cubic

\alpha + \beta + \gamma = -\frac{b}{a}
\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}
\alpha\beta\gamma = -\frac{d}{a}

The proof is the same idea as for quadratics — build the cubic from its roots and match coefficients. Since \alpha, \beta, \gamma are roots, you can write

a(x - \alpha)(x - \beta)(x - \gamma) = ax^3 + bx^2 + cx + d

Expand the left side. Start with (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha\beta. Then multiply by (x - \gamma):

(x^2 - (\alpha + \beta)x + \alpha\beta)(x - \gamma)
= x^3 - \gamma x^2 - (\alpha + \beta)x^2 + (\alpha + \beta)\gamma x + \alpha\beta x - \alpha\beta\gamma
= x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma

Multiply through by a and compare with ax^3 + bx^2 + cx + d:

Notice the alternating signs: the sum of roots gets a minus, the sum of pairwise products is positive, the triple product gets a minus again. This sign pattern — minus, plus, minus — continues for higher-degree equations.

Symmetric functions of roots

The three quantities \alpha + \beta + \gamma, \alpha\beta + \beta\gamma + \gamma\alpha, and \alpha\beta\gamma are called elementary symmetric functions of the roots. They are "symmetric" because swapping any two roots does not change them — \alpha + \beta + \gamma is the same as \beta + \gamma + \alpha.

The power of Vieta's formulas is that you can compute many expressions involving the roots without knowing the roots themselves. You only need the coefficients.

Take x^3 - 6x^2 + 11x - 6 = 0 from the start. Here a = 1, b = -6, c = 11, d = -6. So:

\alpha + \beta + \gamma = 6, \qquad \alpha\beta + \beta\gamma + \gamma\alpha = 11, \qquad \alpha\beta\gamma = 6

(And indeed, the roots are 1, 2, 3: their sum is 6, their pairwise products sum to 2 + 6 + 3 = 11, and their product is 6.)

Now suppose someone asks: what is \alpha^2 + \beta^2 + \gamma^2? You do not need to know the individual roots. Use the identity:

\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)
= 36 - 22 = 14

And 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14. Confirmed.

What about \alpha^3 + \beta^3 + \gamma^3? There is a standard identity:

\alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)^3 - 3(\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) + 3\alpha\beta\gamma
= 216 - 3(6)(11) + 3(6) = 216 - 198 + 18 = 36

Check: 1 + 8 + 27 = 36. These identities are called Newton's power sum formulas, and they let you climb from elementary symmetric functions to any power sum you need.

Vieta's formulas connecting roots 1, 2, 3 to coefficients of x cubed minus 6x squared plus 11x minus 6A diagram showing three roots alpha equals 1, beta equals 2, gamma equals 3 on the left, connected by arrows to three coefficient relationships on the right: sum of roots equals 6, sum of pairwise products equals 11, product of roots equals 6. Roots α = 1 β = 2 γ = 3 From coefficients α+β+γ = 6 αβ+βγ+γα = 11 αβγ = 6 Vieta's formulas read from −b/a, c/a, −d/a
Vieta's formulas for the cubic $x^3 - 6x^2 + 11x - 6 = 0$. The three roots $1, 2, 3$ are connected to the coefficients by the elementary symmetric functions. The sum of roots, sum of pairwise products, and product of roots can all be read directly from the equation without solving it.

Solving cubics: the special cases

Most cubics you encounter in school belong to one of a few types that yield to direct methods.

Case 1: One root is visible

If you can spot a rational root — often by testing small integers — you can factor it out and reduce to a quadratic. The Rational Root Theorem says that any rational root p/q of a polynomial with integer coefficients must have p dividing the constant term and q dividing the leading coefficient. For x^3 - 6x^2 + 11x - 6 = 0, the candidates are \pm 1, \pm 2, \pm 3, \pm 6. Testing x = 1 gives 0, so (x - 1) is a factor. Divide:

x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)

Case 2: The cubic has no x^2 term (depressed cubic)

A cubic of the form x^3 + px + q = 0 — with no x^2 term — is called a depressed cubic. Any cubic can be converted into this form by a substitution. If you start with x^3 + bx^2 + cx + d = 0 (with leading coefficient 1), substitute x = t - b/3. This eliminates the x^2 term, leaving a depressed cubic in t.

For example, take x^3 - 3x^2 + 3x - 1 = 0. Substitute x = t + 1 (since b/3 = 1):

(t+1)^3 - 3(t+1)^2 + 3(t+1) - 1
= t^3 + 3t^2 + 3t + 1 - 3t^2 - 6t - 3 + 3t + 3 - 1 = t^3

So t^3 = 0, giving t = 0 and x = 1 (a triple root). That particular example collapsed entirely, but the substitution works in general.

Case 3: Sum of cubes or difference of cubes

Cubics like x^3 - 8 = 0 or x^3 + 27 = 0 factor using the identities:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)
a^3 + b^3 = (a + b)(a^2 - ab + b^2)

So x^3 - 8 = (x - 2)(x^2 + 2x + 4) = 0. The first factor gives x = 2. The second is a quadratic with discriminant 4 - 16 = -12 < 0, so the other two roots are complex.

Interactive graph of x cubed minus 6x squared plus 11x minus 6A coordinate plane showing the cubic y equals x cubed minus 6x squared plus 11x minus 6. The curve crosses the x-axis at x equals 1, 2, and 3. A draggable red point on the curve shows the current coordinates. x y 1 2 3 4 drag the red point
The cubic $f(x) = x^3 - 6x^2 + 11x - 6$. Drag the red point along the curve and watch it cross zero at $x = 1$, $x = 2$, and $x = 3$ — the three roots predicted by Vieta's formulas. Between consecutive roots, the curve dips below or rises above the axis, creating the characteristic S-shape.

Cardano's method: an introduction

For the general cubic, there is a closed-form solution — but it is far less tidy than the quadratic formula. The method works on a depressed cubic t^3 + pt + q = 0 (any cubic can be reduced to this form, as shown above).

The key idea is to write t = u + v and then choose u and v cleverly. Substituting into t^3 + pt + q = 0:

(u + v)^3 + p(u + v) + q = 0
u^3 + 3u^2v + 3uv^2 + v^3 + pu + pv + q = 0
u^3 + v^3 + (3uv + p)(u + v) + q = 0

Now impose the condition 3uv + p = 0, i.e., uv = -p/3. Then u^3 + v^3 = -q. Combined with u^3 v^3 = -p^3/27, this means u^3 and v^3 are the two roots of the quadratic:

w^2 + qw - \frac{p^3}{27} = 0

Solve this quadratic to find u^3 and v^3, then take cube roots. The result is:

t = \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}}}

This is Cardano's formula. It looks heavy, but the architecture is logical: solve a quadratic in disguise, then take cube roots. The quantity under the square root, \Delta = q^2/4 + p^3/27, plays the role of the discriminant. When \Delta > 0, the cubic has one real root and two complex roots. When \Delta = 0, at least two roots coincide. When \Delta < 0 — and this is the strange part — all three roots are real, but the formula expresses them through cube roots of complex numbers. This case is called the casus irreducibilis, and it baffled mathematicians for centuries: the answer is real, but the path goes through complex numbers.

Cardano's method: reduce to depressed cubic, then solve via a quadratic in disguiseA flow diagram showing three stages: general cubic on the left, depressed cubic in the middle after substitution x equals t minus b over 3a, and the final roots on the right obtained by solving a hidden quadratic and taking cube roots. General cubic ax³+bx²+cx+d = 0 x = t − b/(3a) shift Depressed cubic t³ + pt + q = 0 t = u + v solve Roots cube roots of quadratic solutions
The architecture of Cardano's method. First, shift the variable to remove the $x^2$ term. Then split the new variable as a sum of two unknowns whose cubes satisfy a quadratic equation. Solve the quadratic, take cube roots, and shift back.

Two worked examples

Example 1: Find all roots of $x^3 - 7x + 6 = 0$

This is already a depressed cubic (no x^2 term). Start by hunting for rational roots.

Step 1. The Rational Root Theorem says candidates are \pm 1, \pm 2, \pm 3, \pm 6. Test x = 1: 1 - 7 + 6 = 0. Found one.

Why: since the leading coefficient is 1 and the constant is 6, any rational root must be an integer dividing 6.

Step 2. Factor out (x - 1) by polynomial division. Divide x^3 - 7x + 6 by (x - 1):

x^3 - 7x + 6 = (x - 1)(x^2 + x - 6)

Why: after removing one linear factor, the cubic reduces to a quadratic, which is much easier to handle.

Step 3. Factor the quadratic x^2 + x - 6. Look for two numbers that multiply to -6 and add to 1: that is 3 and -2.

x^2 + x - 6 = (x + 3)(x - 2)

Why: the product 3 \times (-2) = -6 and the sum 3 + (-2) = 1 match the constant and middle coefficient.

Step 4. Combine everything:

x^3 - 7x + 6 = (x - 1)(x + 3)(x - 2) = 0

The three roots are x = 1, x = -3, and x = 2.

Step 5. Verify with Vieta's formulas. Here a = 1, b = 0, c = -7, d = 6.

  • Sum of roots: 1 + (-3) + 2 = 0 = -b/a. Confirmed.
  • Sum of pairwise products: (1)(-3) + (-3)(2) + (2)(1) = -3 - 6 + 2 = -7 = c/a. Confirmed.
  • Product of roots: (1)(-3)(2) = -6 = -d/a. Confirmed.

Result. The roots are x = -3, 1, 2.

Graph of x cubed minus 7x plus 6 with roots at minus 3, 1, and 2A coordinate plane showing the cubic curve y equals x cubed minus 7x plus 6. The curve crosses the x-axis at x equals minus 3, x equals 1, and x equals 2. The S-shape of the curve is visible, with a local maximum between x equals minus 3 and x equals 1, and a local minimum between x equals 1 and x equals 2. x y −3 −2 −1 1 2 3 −3 1 2
The graph of $y = x^3 - 7x + 6$. The three roots at $x = -3$, $x = 1$, and $x = 2$ are exactly the points where the curve crosses the axis. The curve rises to a local maximum between $-3$ and $1$, dips to a local minimum between $1$ and $2$, then rises again — the characteristic S-shape of a cubic with three real roots. The sum of the three roots is $0$, matching the absent $x^2$ term.

The graph confirms all three roots. Notice that the sum of roots is zero — this is because there is no x^2 term (the coefficient b = 0), so \alpha + \beta + \gamma = -b/a = 0. This is a visual signature of a depressed cubic: the roots are balanced around the origin.

Example 2: If the roots of $x^3 - 9x^2 + 26x - 24 = 0$ are $\alpha, \beta, \gamma$, find $\alpha^2 + \beta^2 + \gamma^2$ and $1/\alpha + 1/\beta + 1/\gamma$ without solving the equation

This example uses Vieta's formulas to extract information purely from the coefficients.

Step 1. Read off the elementary symmetric functions. Here a = 1, b = -9, c = 26, d = -24.

\alpha + \beta + \gamma = 9, \qquad \alpha\beta + \beta\gamma + \gamma\alpha = 26, \qquad \alpha\beta\gamma = 24

Why: Vieta's formulas give -b/a = 9, c/a = 26, -d/a = 24 directly from the coefficients.

Step 2. Compute \alpha^2 + \beta^2 + \gamma^2 using the identity (\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha).

\alpha^2 + \beta^2 + \gamma^2 = 9^2 - 2(26) = 81 - 52 = 29

Why: squaring the sum introduces cross terms 2\alpha\beta + 2\beta\gamma + 2\gamma\alpha, which we know. Subtracting them isolates the sum of squares.

Step 3. Compute 1/\alpha + 1/\beta + 1/\gamma by putting the three fractions over a common denominator.

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} = \frac{26}{24} = \frac{13}{12}

Why: the numerator is exactly the sum of pairwise products, and the denominator is the product of all three roots — both known from Vieta's formulas.

Step 4. Verify by finding the actual roots. Testing x = 2: 8 - 36 + 52 - 24 = 0. Testing x = 3: 27 - 81 + 78 - 24 = 0. Testing x = 4: 64 - 144 + 104 - 24 = 0. The roots are 2, 3, 4.

Check: 4 + 9 + 16 = 29 and 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Both confirmed.

Result. \alpha^2 + \beta^2 + \gamma^2 = 29 and 1/\alpha + 1/\beta + 1/\gamma = 13/12.

Symmetric functions of roots 2, 3, 4 computed from coefficientsA summary diagram showing how the three symmetric functions — sum equals 9, pairwise sum equals 26, product equals 24 — combine through algebraic identities to produce the sum of squares (29) and sum of reciprocals (13 over 12). x³ − 9x² + 26x − 24 = 0 roots: 2, 3, 4 Σα = 9 Σαβ = 26 αβγ = 24 α²+β²+γ² = 9² − 2(26) = 29 Σ(1/α) = 26/24 = 13/12 both results obtained without solving the cubic
The power of Vieta's formulas: starting from just the three coefficients, the sum of squares and the sum of reciprocals drop out through standard identities. The actual roots $2, 3, 4$ are never needed — though they confirm the answers when checked.

The diagram captures the essence of symmetric functions: you can build surprisingly many expressions from just three ingredients. Any expression that does not change when you permute the roots can be computed this way.

Common confusions

Going deeper

If you came here to learn the relationship between roots and coefficients of a cubic and how to solve special cases, you have everything you need — you can stop here. The rest is for readers who want to see the discriminant of a cubic and the historical significance of Cardano's formula.

The cubic discriminant

Just as a quadratic has a discriminant b^2 - 4ac that classifies its roots, a cubic x^3 + px + q = 0 has a discriminant:

\Delta = -4p^3 - 27q^2

When \Delta > 0, all three roots are real and distinct. When \Delta = 0, at least two roots coincide. When \Delta < 0, there is one real root and two complex conjugate roots.

For the general cubic ax^3 + bx^2 + cx + d = 0, the discriminant in terms of the original coefficients is:

\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2

This expression is considerably more involved than the quadratic discriminant, but it serves the same purpose: a single number that sorts the cubic into one of three cases.

Why Cardano's formula matters historically

The formula for solving cubics was one of the most dramatic episodes in the history of mathematics. It was the first time in recorded history that a formula was discovered that went genuinely beyond what the ancient mathematicians knew. The quadratic formula had been understood in various forms for thousands of years — by Babylonians, by Indian mathematicians like Brahmagupta, and by many others. But the cubic was new territory.

The formula forced mathematicians to take complex numbers seriously. When all three roots are real but the discriminant q^2/4 + p^3/27 is negative, Cardano's formula involves square roots of negative numbers — and there is no way around it. This was the historical catalyst that pushed complex numbers from being dismissed as "impossible" to being recognised as essential mathematical objects. The real answer passes through the complex plane, and there is no shortcut.

Where this leads next

Cubic equations connect to several deeper topics in algebra and analysis.