Cubic equation

The discriminant of a polynomial is a function of its coefficients that is zero if and only if the polynomial has a multiple root, or, if it is divisible by the square of a non-constant polynomial. In other words, the discriminant is nonzero if and only if the polynomial is square-free.

If r₁, r₂, r₃ are the three roots (not necessarily distinct nor real) of the cubic ${\displaystyle ax^{3}+bx^{2}+cx+d,}$ then the discriminant is

${\displaystyle a^{4}(r_{1}-r_{2})^{2}(r_{1}-r_{3})^{2}(r_{2}-r_{3})^{2}.}$

The discriminant of the depressed cubic ${\displaystyle t^{3}+pt+q}$ is

${\displaystyle -\left(4\,p^{3}+27\,q^{2}\right).}$

The discriminant of the general cubic ${\displaystyle ax^{3}+bx^{2}+cx+d}$ is

${\displaystyle 18\,abcd-4\,b^{3}d+b^{2}c^{2}-4\,ac^{3}-27\,a^{2}d^{2}.}$

It is the product of ${\displaystyle a^{4}}$ and the discriminant of the corresponding depressed cubic. Using the formula relating the general cubic and the associated depressed cubic, this implies that the discriminant of the general cubic can be written as

${\displaystyle {\frac {4(b^{2}-3ac)^{3}-(2b^{3}-9abc+27a^{2}d)^{2}}{27a^{2}}}.}$

It follows that one of these two discriminants is zero if and only if the other is also zero, and, if the coefficients are real, the two discriminants have the same sign. In summary, the same information can be deduced from either one of these two discriminants.

To prove the preceding formulas, one can use Vieta's formulas to express everything as polynomials in r₁, r₂, r₃, and a. The proof then results in the verification of the equality of two polynomials.

If the coefficients of a polynomial are real numbers, and its discriminant ${\displaystyle \Delta }$ is not zero, there are two cases:

• If ${\displaystyle \Delta >0,}$ the cubic has three distinct real roots
• If ${\displaystyle \Delta <0,}$ the cubic has one real root and two non-real complex conjugate roots.

This can be proved as follows. First, if r is a root of a polynomial with real coefficients, then its complex conjugate is also a root. So the non-real roots, if any, occur as pairs of complex conjugate roots. As a cubic polynomial has three roots (not necessarily distinct) by the fundamental theorem of algebra, at least one root must be real.

As stated above, if r₁, r₂, r₃ are the three roots of the cubic ${\displaystyle ax^{3}+bx^{2}+cx+d}$, then the discriminant is

${\displaystyle \Delta =a^{4}(r_{1}-r_{2})^{2}(r_{1}-r_{3})^{2}(r_{2}-r_{3})^{2}}$

If the three roots are real and distinct, the discriminant is a product of positive reals, that is ${\displaystyle \Delta >0.}$

If only one root, say r₁, is real, then r₂ and r₃ are complex conjugates, which implies that r₂ – r₃ is a purely imaginary number, and thus that (r₂ – r₃)² is real and negative. On the other hand, r₁ – r₂ and r₁ – r₃ are complex conjugates, and their product is real and positive.[23] Thus the discriminant is the product of a single negative number and several positive ones. That is ${\displaystyle \Delta <0.}$

If the discriminant of a cubic is zero, the cubic has a multiple root. If furthermore its coefficients are real, then all of its roots are real.

The discriminant of the depressed cubic ${\displaystyle \;t^{3}+pt+q\;}$ is zero if ${\displaystyle 4p^{3}+27q^{2}=0\;.}$ If p is also zero, then p = q = 0 , and 0 is a triple root of the cubic. If ${\displaystyle \;4p^{3}+27q^{2}=0\;,}$ and p ≠ 0 , then the cubic has a simple root

${\displaystyle t_{1}={\frac {\,3q\,}{p}}}$

and a double root

${\displaystyle t_{2}=t_{3}=-{\frac {\,3q\,}{2p}}~.}$

In other words,

${\displaystyle t^{3}+pt+q=\left(t-{\frac {3q}{p}}\right)\left(t+{\frac {\,3q\,}{2p}}\right)^{2}\;.}$

This result can be proved by expanding the latter product or retrieved by solving the rather simple system of equations resulting from Vieta's formulas.

By using the reduction of a depressed cubic, these results can be extended to the general cubic. This gives: If the discriminant of the cubic ${\displaystyle \;ax^{3}+bx^{2}+cx+d\;}$ is zero, then

• either, if ${\displaystyle b^{2}=3ac\;,}$ the cubic has a triple root

${\displaystyle x_{1}=x_{2}=x_{3}=-{\frac {b}{\,3a\,}}~,}$

and

${\displaystyle ax^{3}+bx^{2}+cx+d=a\left(x+{\frac {b}{\,3a\,}}\right)^{3}}$

• or, if ${\displaystyle \;b^{2}\neq 3ac\;,}$ the cubic has a double root

${\displaystyle x_{2}=x_{3}={\frac {9ad-bc}{\,2(b^{2}-3ac)\,}}~,}$

and a simple root,

${\displaystyle x_{1}={\frac {\,4abc-9a^{2}d-b^{3}\,}{a(b^{2}-3ac)}}~.}$

and thus

${\displaystyle ax^{3}+bx^{2}+cx+d=a\,\left(x-x_{1}\right)\left(x-x_{2}\right)^{2}~.}$

The above results are valid when the coefficients belong to a field of characteristic other than 2 or 3, but must be modified for characteristic 2 or 3, because of the involved divisions by 2 and 3.

The reduction to a depressed cubic works for characteristic 2, but not for characteristic 3. However, in both cases, it is simpler to establish and state the results for the general cubic. The main tool for that is the fact that a multiple root is a common root of the polynomial and its formal derivative. In these characteristics, if the derivative is not a constant, it is a linear polynomial in characteristic 3, and is the square of a linear polynomial in characteristic 2. Therefore, for either characteristic 2 or 3, the derivative has only one root. This allows computing the multiple root, and the third root can be deduced from the sum of the roots, which is provided by Vieta's formulas.

A difference with other characteristics is that, in characteristic 2, the formula for a double root involves a square root, and, in characteristic 3, the formula for a triple root involves a cube root.

When a cubic equation with real coefficients has three real roots, the formulas expressing these roots in terms of radicals involve complex numbers. Galois theory allows proving that when the three roots are real, and none is rational (casus irreducibilis), one cannot express the roots in terms of real radicals. Nevertheless, purely real expressions of the solutions may be obtained using trigonometric functions, specifically in terms of cosines and arccosines.[24] More precisely, the roots of the depressed cubic

${\displaystyle t^{3}+pt+q=0}$

are[25]

${\displaystyle t_{k}=2\,{\sqrt {-{\frac {p}{3}}}}\,\cos \left[\,{\frac {1}{3}}\arccos \left({\frac {3q}{2p}}{\sqrt {\frac {-3}{p}}}\,\right)-{\frac {2\pi k}{3}}\,\right]\qquad {\text{for }}k=0,1,2.}$

This formula is due to François Viète.[22] It is purely real when the equation has three real roots (that is ${\displaystyle 4p^{3}+27q^{2}<0}$). Otherwise, it is still correct but involves complex cosines and arccosines when there is only one real root, and it is nonsensical (division by zero) when p = 0.

This formula can be straightforwardly transformed into a formula for the roots of a general cubic equation, using the back-substitution described in § Depressed cubic.

The formula can be proved as follows: Starting from the equation t³ + pt + q = 0, let us set  t = u cos θ. The idea is to choose u to make the equation coincide with the identity

${\displaystyle 4\cos ^{3}\theta -3\cos \theta -\cos(3\theta )=0.}$

For this, choose ${\displaystyle u=2\,{\sqrt {-{\frac {p}{3}}}}\,,}$ and divide the equation by ${\displaystyle {\frac {u^{3}}{4}}.}$ This gives

${\displaystyle 4\cos ^{3}\theta -3\cos \theta -{\frac {3q}{2p}}\,{\sqrt {\frac {-3}{p}}}=0.}$

Combining with the above identity, one gets

${\displaystyle \cos(3\theta )={\frac {3q}{2p}}{\sqrt {\frac {-3}{p}}}\,,}$

and the roots are thus

${\displaystyle t_{k}=2\,{\sqrt {-{\frac {p}{3}}}}\,\cos \left[{\frac {1}{3}}\arccos \left({\frac {3q}{2p}}{\sqrt {\frac {-3}{p}}}\right)-{\frac {2\pi k}{3}}\right]\qquad {\text{for }}k=0,1,2.}$

When there is only one real root (and p ≠ 0), this root can be similarly represented using hyperbolic functions, as[26][27]

${\displaystyle {\begin{aligned}t_{0}&=-2{\frac {|q|}{q}}{\sqrt {-{\frac {p}{3}}}}\cosh \left[{\frac {1}{3}}\operatorname {arcosh} \left({\frac {-3|q|}{2p}}{\sqrt {\frac {-3}{p}}}\right)\right]\qquad {\text{if }}~4p^{3}+27q^{2}>0~{\text{ and }}~p<0,\\t_{0}&=-2{\sqrt {\frac {p}{3}}}\sinh \left[{\frac {1}{3}}\operatorname {arsinh} \left({\frac {3q}{2p}}{\sqrt {\frac {3}{p}}}\right)\right]\qquad {\text{if }}~p>0.\end{aligned}}}$

If p ≠ 0 and the inequalities on the right are not satisfied (the case of three real roots), the formulas remain valid but involve complex quantities.

When p = ±3, the above values of t₀ are sometimes called the Chebyshev cube root.[28] More precisely, the values involving cosines and hyperbolic cosines define, when p = −3, the same analytic function denoted C_1/3(q), which is the proper Chebyshev cube root. The value involving hyperbolic sines is similarly denoted S_1/3(q), when p = 3.

Omar Khayyám's geometric solution of a cubic equation, for the case m = 2, n = 16, giving the root 2. The intersection of the vertical line on the x-axis at the center of the circle is happenstance of the example illustrated.

For solving the cubic equation x³ + m²x = n where n > 0, Omar Khayyám constructed the parabola y = x²/m, the circle that has as a diameter the line segment [0, n/m²] on the positive x-axis, and a vertical line through the point where the circle and the parabola intersect above the x-axis. The solution is given by the length of the horizontal line segment from the origin to the intersection of the vertical line and the x-axis (see the figure).

A simple modern proof is as follows. Multiplying the equation by x/m² and regrouping the terms gives

${\displaystyle {\frac {x^{4}}{m^{2}}}=x\left({\frac {n}{m^{2}}}-x\right).}$

The left-hand side is the value of y² on the parabola. The equation of the circle being y² + x(x − n/m²) = 0, the right hand side is the value of y² on the circle.

A cubic equation with real coefficients can be solved geometrically using compass, straightedge, and an angle trisector if and only if it has three real roots.[29]^{: Thm. 1}

A cubic equation can be solved by compass-and-straightedge construction (without trisector) if and only if it has a rational root. This implies that the old problems of angle trisection and doubling the cube, set by ancient Greek mathematicians, cannot be solved by compass-and-straightedge construction.

For the cubic (1) with three real roots, the roots are the projection on the x-axis of the vertices A, B, and C of an equilateral triangle. The center of the triangle has the same x-coordinate as the inflection point.

Viète's trigonometric expression of the roots in the three-real-roots case lends itself to a geometric interpretation in terms of a circle.[22][30] When the cubic is written in depressed form (2), t³ + pt + q = 0, as shown above, the solution can be expressed as

${\displaystyle t_{k}=2{\sqrt {-{\frac {p}{3}}}}\cos \left({\frac {1}{3}}\arccos \left({\frac {3q}{2p}}{\sqrt {\frac {-3}{p}}}\right)-k{\frac {2\pi }{3}}\right)\quad {\text{for}}\quad k=0,1,2\,.}$

Here ${\displaystyle \arccos \left({\frac {3q}{2p}}{\sqrt {\frac {-3}{p}}}\right)}$ is an angle in the unit circle; taking 1/3 of that angle corresponds to taking a cube root of a complex number; adding −k2π/3 for k = 1, 2 finds the other cube roots; and multiplying the cosines of these resulting angles by ${\displaystyle 2{\sqrt {-{\frac {p}{3}}}}}$ corrects for scale.

For the non-depressed case (1) (shown in the accompanying graph), the depressed case as indicated previously is obtained by defining t such that x = t − b/3a so t = x + b/3a. Graphically this corresponds to simply shifting the graph horizontally when changing between the variables t and x, without changing the angle relationships. This shift moves the point of inflection and the centre of the circle onto the y-axis. Consequently, the roots of the equation in t sum to zero.

The slope of line RA is twice that of RH. Denoting the complex roots of the cubic as g ± hi, g = OM (negative here) and h = √tan ORH = √slope of line RH = BE = DA.

When the graph of a cubic function is plotted in the Cartesian plane, if there is only one real root, it is the abscissa (x-coordinate) of the horizontal intercept of the curve (point R on the figure). Further,[31][32][33] if the complex conjugate roots are written as g ± hi, then the real part g is the abscissa of the tangency point H of the tangent line to cubic that passes through x-intercept R of the cubic (that is the signed length RM, negative on the figure). The imaginary parts ±h are the square roots of the tangent of the angle between this tangent line and the horizontal axis.

With one real and two complex roots, the three roots can be represented as points in the complex plane, as can the two roots of the cubic's derivative. There is an interesting geometrical relationship among all these roots.

The points in the complex plane representing the three roots serve as the vertices of an isosceles triangle. (The triangle is isosceles because one root is on the horizontal (real) axis and the other two roots, being complex conjugates, appear symmetrically above and below the real axis.) Marden's theorem says that the points representing the roots of the derivative of the cubic are the foci of the Steiner inellipse of the triangle—the unique ellipse that is tangent to the triangle at the midpoints of its sides. If the angle at the vertex on the real axis is less than π/3 then the major axis of the ellipse lies on the real axis, as do its foci and hence the roots of the derivative. If that angle is greater than π/3, the major axis is vertical and its foci, the roots of the derivative, are complex conjugates. And if that angle is π/3, the triangle is equilateral, the Steiner inellipse is simply the triangle's incircle, its foci coincide with each other at the incenter, which lies on the real axis, and hence the derivative has duplicate real roots.

This method is due to Scipione del Ferro and Tartaglia, but is named after Gerolamo Cardano who first published it in his book Ars Magna (1545).

This method applies to a depressed cubic t³ + pt + q = 0. The idea is to introduce two variables u and v such that u + v = t and to substitute this in the depressed cubic, giving

${\displaystyle u^{3}+v^{3}+(3uv+p)(u+v)+q=0.}$

At this point Cardano imposed the condition 3uv + p = 0. This removes the third term in previous equality, leading to the system of equations

${\displaystyle {\begin{aligned}u^{3}+v^{3}&=-q\\uv&=-{\frac {p}{3}}.\end{aligned}}}$

Knowing the sum and the product of u³ and v³, one deduces that they are the two solutions of the quadratic equation

${\displaystyle (x-u^{3})(x-v^{3})=x^{2}-(u^{3}+v^{3})x+u^{3}v^{3}=x^{2}-(u^{3}+v^{3})x+(uv)^{3}=0,}$

so

${\displaystyle x^{2}+qx-{\frac {p^{3}}{27}}=0,}$

The discriminant of this equation is ${\displaystyle \Delta =q^{2}+{\frac {4p^{3}}{27}}}$, and assuming it is positive, real solutions to this equations are (after folding division by 4 under the square root):

${\displaystyle -{\frac {q}{2}}\pm {\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}.}$

So (without loss of generality in choosing u or v):

${\displaystyle u={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}.}$

${\displaystyle v={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}.}$

As u + v = t, the sum of the cube roots of these solutions is a root of the equation. That is

${\displaystyle t={\sqrt[{3}]{-{q \over 2}+{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}+{\sqrt[{3}]{-{q \over 2}-{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}}$

is a root of the equation; this is Cardano's formula.

This works well when ${\displaystyle 4p^{3}+27q^{2}>0,}$ but, if ${\displaystyle 4p^{3}+27q^{2}<0,}$ the square root appearing in the formula is not real. As a complex number has three cube roots, using Cardano's formula without care would provide nine roots, while a cubic equation cannot have more than three roots. This was clarified first by Rafael Bombelli in his book L'Algebra (1572). The solution is to use the fact that uv = –p/3, that is v = –p/3u. This means that only one cube root needs to be computed, and leads to the second formula given in § Cardano's formula.

The other roots of the equation can be obtained by changing of cube root, or, equivalently, by multiplying the cube root by each of the two primitive cube roots of unity, which are ${\displaystyle {\frac {-1\pm {\sqrt {-3}}}{2}}.}$

Vieta's substitution is a method introduced by François Viète (Vieta is his Latin name) in a text published posthumously in 1615, which provides directly the second formula of § Cardano's method, and avoids the problem of computing two different cube roots.[34]

Starting from the depressed cubic t³ + pt + q = 0, Vieta's substitution is t = w – p/3w.[35]

The substitution t = w – p/3w transforms the depressed cubic into

${\displaystyle w^{3}+q-{\frac {p^{3}}{27w^{3}}}=0.}$

Multiplying by w³, one gets a quadratic equation in w³:

${\displaystyle (w^{3})^{2}+q(w^{3})-{\frac {p^{3}}{27}}=0.}$

Let

${\displaystyle W=-{\frac {q}{2}}\pm {\sqrt {{\frac {p^{3}}{27}}+{\frac {q^{2}}{4}}}}}$

be any nonzero root of this quadratic equation. If w₁, w₂ and w₃ are the three cube roots of W, then the roots of the original depressed cubic are w₁ − p/3w₁, w₂ − p/3w₂, and w₃ − p/3w₃. The other root of the quadratic equation is ${\displaystyle \textstyle -{\frac {p^{3}}{27W}}.}$ This implies that changing the sign of the square root exchanges w_i and − p/3w_i for i = 1, 2, 3, and therefore does not change the roots. This method only fails when both roots of the quadratic equation are zero, that is when p = q = 0, in which case the only root of the depressed cubic is 0.

In his paper Réflexions sur la résolution algébrique des équations ("Thoughts on the algebraic solving of equations"),[36] Joseph Louis Lagrange introduced a new method to solve equations of low degree in a uniform way, with the hope that he could generalize it for higher degrees. This method works well for cubic and quartic equations, but Lagrange did not succeed in applying it to a quintic equation, because it requires solving a resolvent polynomial of degree at least six.[37][38][39] Except that nobody succeeded before to solve the problem, this was the first indication of the non-existence of an algebraic formula for degrees 5 and higher; this was later proved as the Abel–Ruffini theorem. Nevertheless, the modern methods for solving solvable quintic equations are mainly based on Lagrange's method.[39]

In the case of cubic equations, Lagrange's method gives the same solution as Cardano's. Lagrange's method can be applied directly to the general cubic equation ax³ + bx² + cx + d = 0, but the computation is simpler with the depressed cubic equation, t³ + pt + q = 0.

Lagrange's main idea was to work with the discrete Fourier transform of the roots instead of with the roots themselves. More precisely, let ξ be a primitive third root of unity, that is a number such that ξ³ = 1 and ξ² + ξ + 1 = 0 (when working in the space of complex numbers, one has ${\displaystyle \textstyle \xi ={\frac {-1\pm i{\sqrt {3}}}{2}}=e^{2i\pi /3},}$ but this complex interpretation is not used here). Denoting x₀, x₁ and x₂ the three roots of the cubic equation to be solved, let

${\displaystyle {\begin{aligned}s_{0}&=x_{0}+x_{1}+x_{2},\\s_{1}&=x_{0}+\xi x_{1}+\xi ^{2}x_{2},\\s_{2}&=x_{0}+\xi ^{2}x_{1}+\xi x_{2},\end{aligned}}}$

be the discrete Fourier transform of the roots. If s₀, s₁ and s₂ are known, the roots may be recovered from them with the inverse Fourier transform consisting of inverting this linear transformation; that is,

${\displaystyle {\begin{aligned}x_{0}&={\tfrac {1}{3}}(s_{0}+s_{1}+s_{2}),\\x_{1}&={\tfrac {1}{3}}(s_{0}+\xi ^{2}s_{1}+\xi s_{2}),\\x_{2}&={\tfrac {1}{3}}(s_{0}+\xi s_{1}+\xi ^{2}s_{2}).\end{aligned}}}$

By Vieta's formulas, s₀ is known to be zero in the case of a depressed cubic, and −b/a for the general cubic. So, only s₁ and s₂ need to be computed. They are not symmetric functions of the roots (exchanging x₁ and x₂ exchanges also s₁ and s₂), but some simple symmetric functions of s₁ and s₂ are also symmetric in the roots of the cubic equation to be solved. Thus these symmetric functions can be expressed in terms of the (known) coefficients of the original cubic, and this allows eventually expressing the s_i as roots of a polynomial with known coefficients. This works well for every degree, but, in degrees higher than four, the resulting polynomial that has the s_i as roots has a degree higher than that of the initial polynomial, and is therefore unhelpful for solving. This is the reason for which Lagrange's method fails in degrees five and higher.

In the case of a cubic equation, ${\displaystyle P=s_{1}s_{2},}$ and ${\displaystyle S=s_{1}^{3}+s_{2}^{3}}$ are such symmetric polynomials (see below). It follows that ${\displaystyle s_{1}^{3}}$ and ${\displaystyle s_{2}^{3}}$ are the two roots of the quadratic equation ${\displaystyle z^{2}-Sz+P^{3}=0.}$ Thus the resolution of the equation may be finished exactly as with Cardano's method, with ${\displaystyle s_{1}}$ and ${\displaystyle s_{2}}$ in place of u and v.

In the case of the depressed cubic, one has ${\displaystyle x_{0}={\tfrac {1}{3}}(s_{1}+s_{2})}$ and ${\displaystyle s_{1}s_{2}=-3p,}$ while in Cardano's method we have set ${\displaystyle x_{0}=u+v}$ and ${\displaystyle uv={\tfrac {1}{3}}p.}$ Thus, up to the exchange of u and v, we have ${\displaystyle s_{1}=3u}$ and ${\displaystyle s_{2}=3v.}$ In other words, in this case, Cardano's method and Lagrange's method compute exactly the same things, up to a factor of three in the auxiliary variables, the main difference being that Lagrange's method explains why these auxiliary variables appear in the problem.

A straightforward computation using the relations ξ³ = 1 and ξ² + ξ + 1 = 0 gives

${\displaystyle {\begin{aligned}P&=s_{1}s_{2}=x_{0}^{2}+x_{1}^{2}+x_{2}^{2}-(x_{0}x_{1}+x_{1}x_{2}+x_{2}x_{0}),\\S&=s_{1}^{3}+s_{2}^{3}=2(x_{0}^{3}+x_{1}^{3}+x_{2}^{3})-3(x_{0}^{2}x_{1}+x_{1}^{2}x_{2}+x_{2}^{2}x_{0}+x_{0}x_{1}^{2}+x_{1}x_{2}^{2}+x_{2}x_{0}^{2})+12x_{0}x_{1}x_{2}.\end{aligned}}}$

This shows that P and S are symmetric functions of the roots. Using Newton's identities, it is straightforward to express them in terms of the elementary symmetric functions of the roots, giving

${\displaystyle {\begin{aligned}P&=e_{1}^{2}-3e_{2},\\S&=2e_{1}^{3}-9e_{1}e_{2}+27e_{3},\end{aligned}}}$

with e₁ = 0, e₂ = p and e₃ = −q in the case of a depressed cubic, and e₁ = −b/a, e₂ = c/a and e₃ = −d/a, in the general case.

• Angle trisection and doubling the cube are two ancient problems of geometry that have been proved to not be solvable by straightedge and compass construction, because they are equivalent to solving a cubic equation.
• Marden's theorem states that the foci of the Steiner inellipse of any triangle can be found by using the cubic function whose roots are the coordinates in the complex plane of the triangle's three vertices. The roots of the first derivative of this cubic are the complex coordinates of those foci.
• The area of a regular heptagon can be expressed in terms of the roots of a cubic. Further, the ratios of the long diagonal to the side, the side to the short diagonal, and the negative of the short diagonal to the long diagonal all satisfy a particular cubic equation. In addition, the ratio of the inradius to the circumradius of a heptagonal triangle is one of the solutions of a cubic equation. The values of trigonometric functions of angles related to ${\displaystyle 2\pi /7}$ satisfy cubic equations.
• Given the cosine (or other trigonometric function) of an arbitrary angle, the cosine of one-third of that angle is one of the roots of a cubic.
• The solution of the general quartic equation relies on the solution of its resolvent cubic.
• The eigenvalues of a 3×3 matrix are the roots of a cubic polynomial which is the characteristic polynomial of the matrix.
• The characteristic equation of a third-order constant coefficients or Cauchy–Euler (equidimensional variable coefficients) linear differential equation or difference equation is a cubic equation.
• Intersection points of cubic Bézier curve and straight line can be computed using direct cubic equation representing Bézier curve.
• Critical points of a quartic function are found by solving a cubic equation (the derivative set equal to zero).
• Inflection points of a quintic function are the solution of a cubic equation (the second derivative set equal to zero).

• In analytical chemistry, the Charlot equation, which can be used to find the pH of buffer solutions, can be solved using a cubic equation.
• In thermodynamics, equations of state (which relate pressure, volume, and temperature of a substances) are cubic in the volume.
• Kinematic equations involving linear rates of acceleration are cubic.
• The speed of seismic Rayleigh waves is a solution of the Rayleigh wave cubic equation.
• The steady state speed of a vehicle moving on a slope with air friction for a given input power is solved by a depressed cubic equation.