Math Detours · Grade 11 · Around the corner · 5

The cubic formula, and the numbers it forced into existence

Solve \(x^3 = 15x + 4\) by Cardano's method and the answer, 4, arrives wrapped in square roots of negative numbers. This is why complex numbers were invented.

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Tonight's idea

The quadratic formula has a cousin for cubics. Take a cubic with no \(x^2\) term, written \(x^3 = px + q\). The trick is to look for the answer as a sum, \(x = u + v\). Expand the cube and regroup:

\[(u + v)^3 = 3uv(u + v) + (u^3 + v^3).\]

This matches \(x^3 = px + q\) if \(3uv = p\) and \(u^3 + v^3 = q\). So \(u^3\) and \(v^3\) are two numbers with sum \(q\) and product \((p/3)^3\). Two numbers with a known sum and product are the roots of a quadratic (grade 10): \(t^2 - qt + (p/3)^3 = 0\). Solve it, take cube roots, and add.

Cardano's formula. \(x = \sqrt[3]{\tfrac{q}{2} + \sqrt{D}} + \sqrt[3]{\tfrac{q}{2} - \sqrt{D}}\), where \(D = \left(\tfrac{q}{2}\right)^2 - \left(\tfrac{p}{3}\right)^3\).

The Italian mathematician Gerolamo Cardano published the method in 1545. It has a strange flaw. When \(D\) is negative the formula asks for the square root of a negative number, and that is exactly when the cubic has three different real roots.

−2 − √3 −2 + √3 4 y = x³ − 15x − 4x
Three real roots, yet the formula reaches 4 only through \(\sqrt{-121}\).

Rafael Bombelli, later that century, calculated with those square roots anyway, by the ordinary rules. They cancelled and a real root came out. Quadratics never forced anyone to accept \(\sqrt{-1}\). The cubic did.

Watch one: \(x^3 = 15x + 4\)

Here \(\tfrac{p}{3} = 5\) and \(\tfrac{q}{2} = 2\), so \(D = 2^2 - 5^3 = -121\) and \(\sqrt{D} = 11i\), where \(i^2 = -1\). The formula gives

\[x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}.\]

Now cube \(2 + i\):

\[(2 + i)^3 = 8 + 12i + 6i^2 + i^3 = 8 + 12i - 6 - i = 2 + 11i.\]

So \(2 + i\) is a cube root of \(2 + 11i\), and in the same way \(2 - i\) is a cube root of \(2 - 11i\). Then \(x = (2 + i) + (2 - i) = 4\). The imaginary parts cancel. Check: \(4^3 = 64 = 15 \times 4 + 4\).

Drag a complex number around and see what multiplying by \(i\) and squaring do to it: mathdetours.com/demos/complex.html.

Your turn

  1. Use Cardano's formula to solve \(x^3 = 6x + 9\). Here \(D\) is positive and both cube roots are whole numbers. Check your answer in the equation.
  2. Write \(x^3 + 3x = 4\) as \(x^3 = px + q\) and apply the formula. Then show that \(\left(\tfrac{1 + \sqrt5}{2}\right)^3 = 2 + \sqrt5\), and use it to simplify your answer to a whole number.
  3. Show that \((3 + i)^3 = 18 + 26i\). Use this to solve \(x^3 = 30x + 36\) by the formula.
  4. The formula needs a cubic with no squared term. Substitute \(y = x + 2\) into \(y^3 - 6y^2 - 3y + 18 = 0\) and show that it becomes \(x^3 = 15x + 4\). You know \(x = 4\) is one root. Divide by \(x - 4\) to find the other two, and so give all three values of \(y\).

Stretch: find the cube root yourself

Solve \(x^3 = 6x + 4\) by the formula. You will need a cube root of \(2 + 2i\): try \(a + bi\) with \(a\) and \(b\) each 1 or \(-1\). Then find the other two roots of the cubic.

Answers

  1. \(\tfrac{p}{3} = 2\), \(\tfrac{q}{2} = \tfrac92\), so \(D = \tfrac{81}{4} - 8 = \tfrac{49}{4}\) and \(\sqrt{D} = \tfrac72\). Then \(x = \sqrt[3]{8} + \sqrt[3]{1} = 2 + 1 = 3\). Check: \(27 = 18 + 9\).
  2. \(x^3 = -3x + 4\), so \(\tfrac{p}{3} = -1\), \(\tfrac{q}{2} = 2\), \(D = 4 + 1 = 5\), and \(x = \sqrt[3]{2 + \sqrt5} + \sqrt[3]{2 - \sqrt5}\). Cubing, \(\left(\tfrac{1 + \sqrt5}{2}\right)^3 = \tfrac{1 + 3\sqrt5 + 15 + 5\sqrt5}{8} = 2 + \sqrt5\), and in the same way \(\left(\tfrac{1 - \sqrt5}{2}\right)^3 = 2 - \sqrt5\). So \(x = \tfrac{1 + \sqrt5}{2} + \tfrac{1 - \sqrt5}{2} = 1\). Check: \(1 + 3 = 4\).
  3. \((3 + i)^3 = 27 + 27i + 9i^2 + i^3 = 27 + 27i - 9 - i = 18 + 26i\). For the cubic, \(\tfrac{p}{3} = 10\), \(\tfrac{q}{2} = 18\), \(D = 324 - 1000 = -676\), \(\sqrt{D} = 26i\). So \(x = \sqrt[3]{18 + 26i} + \sqrt[3]{18 - 26i} = (3 + i) + (3 - i) = 6\). Check: \(216 = 180 + 36\).
  4. \((x+2)^3 - 6(x+2)^2 - 3(x+2) + 18 = x^3 - 15x - 4\): the \(x^2\) terms are \(6x^2 - 6x^2 = 0\), the \(x\) terms \(12x - 24x - 3x = -15x\), the constants \(8 - 24 - 6 + 18 = -4\). Then \(x^3 - 15x - 4 = (x - 4)(x^2 + 4x + 1)\), so \(x = 4\) or \(x = -2 \pm \sqrt3\). Adding 2: \(y = 6\), \(\sqrt3\) or \(-\sqrt3\).

Stretch. \(\tfrac{p}{3} = 2\), \(\tfrac{q}{2} = 2\), \(D = 4 - 8 = -4\), \(\sqrt{D} = 2i\). Since \((-1 + i)^2 = -2i\), \((-1 + i)^3 = -2i(-1 + i) = 2 + 2i\). So \(x = (-1 + i) + (-1 - i) = -2\). Check: \(-8 = -12 + 4\). Then \(x^3 - 6x - 4 = (x + 2)(x^2 - 2x - 2)\), giving \(x = 1 \pm \sqrt3\). Again three real roots, reached through \(i\).

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