Cardano's Solution

of the

Depressed Cubic Equation

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    In 1545 Gerolamo Cardano (1501 Sep 24 - 1576 Sep 20) published Artis Magnae, sive de regulis algebraicis (Of the Great Art, or On the Rules of Algebra), in which he published, among other things, a solution of the depressed cubic equation. In modern notation that equation looks like this:

(Eq'n 1)

Today we would have a difficult time finding a solution of that equation, but Cardano and his contemporaries had an advantage that we have all but lost: the algebraic notation and rules for using it that are familiar to us were yet poorly developed in the Sixteenth Century, so mathematicians of the time still relied heavily on geometry and geometric analogies in working out their problems. Using just such an analogy makes the solution of Equation 1 seem almost trivial.

    Imagine a cube whose sides have a length p. Imagine further that three planes oriented parallel to the cube's faces cut the cube at a distance of q (q<<p) from the cube's top face, its rear face, and its right face. Those planes thus cut the cube into eight chunks whose volumes we can calculate readily: we have,

    1. a large cube, (p-q)3;

    2. a small cube, q3;

    3. three slabs, 3x(p-q)2q; and

    4. three bars, 3x(p-q)q2.

We now have two ways in which we can represent the volume of the cube and we can equate them to each other to get

(Eq'n 2)

If we had defined r=p-q and then set out to calculate p3=(r+q)3, we could have obtained Equation 2 through direct use of the binomial theorem, but Blaise Pascal didn't give that theorem to European mathematicians until a century after Cardano's time.

    Starting by subtracting the cube of q from both sides, we can rearrange Equation 2 to get

(Eq'n 3)

That equation corresponds to Equation 1 with

(Eq'n 4)

(Eq'n 5)

and

(Eq'n 6)

    That seems like a bit of trouble to go to just to make the problem more complex. But in fact it gives us a handle on Equation 1 that lets us solve it. We know that we can represent any number as the difference between two other numbers. Once we have made that representation, Equation 1 then determines the relationships between those numbers and its coefficients. If we can solve those relationships for p and q in terms of a and b, then we have our solution.

    We have immediately from Equation 5 that

(Eq'n 7)

That lets us eliminate q and rewrite Equation 6 as

(Eq'n 8)

If we multiply that equation by p3 and rearrange the terms, we get

(Eq'n 9)

which is quadratic in p3. We already know how to solve a quadratic equation, so we have

(Eq'n 10)

We want real-valued solutions, so we take the positive square root in that equation and get

(Eq'n 11)

From Equation 6 we have

(Eq'n 12)

so we get

(Eq'n 13)

And finally, by way of Equation 4, we get the complete solution,

(Eq'n 14)

    Now I want to solve a depressed cubic equation with a negative coefficient; that is,

(Eq'n 15)

In this case we begin by using the binomial theorem to calculate the cube of the sum of p and q;

(Eq'n 16)

Subtracting from both sides the middle two terms on the right gives us

(Eq'n 17)

which has the same form as does Equation 15. Again, making the analogy between Equations 15 and 17, we have

(Eq'n 18)

(Eq'n 19)

and

(Eq'n 20)

We take q=a/3p and substitute it into Equation 20, which we then multiply by p3 to get the quadratic equation

(Eq'n 21)

That readily yields the solution

(Eq'n 22)

Substituting that result into Equation 20 then lets us derive

(Eq'n 23)

for a final solution of

(Eq'n 24)

    That solution will test our faith in mathematics. If we have a situation in which a3/27 is larger than b2/4, the square root will be an imaginary number and our solution will be the sum of the cube roots of two complex numbers. If we require a real-valued solution, will we get it? Here's one way to find out.

Example Application: TerraJove

    Some time in the next thousand years or so our descendants may decide to expand the amount of Earth-like land area available to them. The most expedient way of achieving that expansion involves building a ring around Jupiter. The ring must rotate once every twenty-four hours, to give its inhabitants a normal day-night cycle (huge Fresnel lenses in deep space will concentrate the sun's light to provide normal illumination on the ring's outer surface.). And a person standing on the outer surface must feel normal Earth gravity directed toward Jupiter. On what radius must the builders construct the ring?

    We need to find the radial distance R from Jupiter's center at which the acceleration due to Jupiter's gravity exceeds the acceleration due to the centrifugal force on the ring by g=9.8 meters per second per second. We thus have,

(Eq'n 25)

in which we have Jupiter's gravitational constant (MgG=1.266x1017 meters cubed per second squared) and the angular velocity of the ring (ω=7.27x10-5 radian per second; note that this differs from the 7.29x10-5 radian per second for Earth because Jupiter's motion on its orbit does not add almost one degree per day to the standard 360 as Earth's orbital motion does.). If we divide Equation 25 by MgG and R, we get

(Eq'n 26)

which is the depressed cubic equation that we want to solve for x=1/R. We have

(Eq'n 27)

and

(Eq'n 28)

So we have

(Eq'n 29)

and

(Eq'n 30)

which gives us

(Eq'n 31)

That result tells us that in order to complete the solution we must calculate the cube roots of two complex numbers.

    In calculating the powers and roots of a complex number Z, we first express the number in its polar form,

(Eq'n 32)

in which A (a real-valued number) expresses the length of the line extending from the origin of the complex plane to the point in that plane representing Z and θ represents the angle that line makes with the positive real axis. If we divide Z into its real and imaginary parts, we have

(Eq'n 33)

and thence

(Eq'n 34)

and

(Eq'n 35)

In our current calculation, then, we have

(Eq'n 36)

and

(Eq'n 37)

    In light of Equation 32 we have Equation 22 as

(Eq'n 38)

and Equation 23 as

(Eq'n 39)

We then have directly

(Eq'n 40)

Note that in summing the exponentials we have automatically canceled out the imaginary parts of the solution to obtain an entirely real-valued solution, as we require. Plugging in the relevant numbers and turning the arithmetic crank gives us

(Eq'n 41)

That, in turn, gives us

(Eq'n 42)

Thus we know that our ring around Jupiter must lie 110,438 kilometers from the planet's center. From that datum we can calculate other facts that we need to know about the ring, such as how fast it moves in its rotation about the planet (8.029 kilometers per second).

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