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Solving cubics

One idea for solving cubic equations is as follows. Suppose we take an expression

\[ \sqrt[3]{A} + \sqrt[3]{B} \]

and cube it. We obtain

\begin{align*} \left(\sqrt[3]{A} + \sqrt[3]{B}\right)^3 &= A + 3\left( \sqrt[3]{A} \right)^2 \sqrt[3]{B} + 3 \sqrt[3]{A} \left( \sqrt[3]{B} \right)^2 + B \\ &= A + B + 3 \sqrt[3]{AB} \left( \sqrt[3]{A} + \sqrt[3]{B} \right). \end{align*}

This says that \(\sqrt[3]{A} + \sqrt[3]{B}\) is a solution of the equation

\[ x^3 = (A+B) + 3\sqrt[3]{AB} \; x. \]

So, given a cubic equation \(x^3 +ax+b=0\), we can find a solution as follows:

  1. rewrite the equation in the form \(x^3 = p+qx\)
  2. find \(A\) and \(B\) so that \(A+B = p\) and \(3\sqrt[3]{AB} = q\)
  3. a solution is then \(\sqrt[3]{A} + \sqrt[3]{B}\).

Example

Find a solution to the equation \(x^3-18x-30 = 0\).

Solution

Rewriting the equation as \(x^3 = 30 + 18x\), we must find \(A\) and \(B\) such that \(A+B=30\) and \(3 \sqrt[3]{AB} = 18\), i.e., \(AB = 216\). Substituting \(B=\frac{216}{A}\) into \(A+B=30\) gives

\[ A + \frac{216}{A} = 30, \qquad \text{which is equivalent to} \quad A^2 - 30A + 216 = 0. \]

This quadratic has solutions \(A=12,18\). So we obtain a solution \(A=12\), \(B=18\), and a solution to the original cubic is then

\[ x = \sqrt[3]{12} + \sqrt[3]{18}. \]

We might note that this method only gives us one of the roots. However, at least in principle, knowing one root, we can factorise our cubic into linear and quadratic factors, and proceed. (In practice however, this method can lead to severe algebraic complications when there are three real roots, and complex numbers appear.)

Unfortunately, the above method only works when the cubic equation has no \(x^2\) term. However, you can always make a substitution to get rid of the \(x^2\) term!

For instance, suppose we want to solve the cubic equation

\[ x^3 + 3x^2 - 15x - 47 = 0. \]

Let \(z=x+1\); then the above equation can be written in terms of \(z\):

\[ (z-1)^3 + 3(z-1)^2 - 15(z-1) - 47 = z^3 - 18z - 30 = 0. \]

So our equation reduces to the previous example. In general, given a cubic equation

\[ x^3 + ax^2 + bx + c = 0, \]

letting

\[ z = x + \dfrac{a}{3} \]

and rewriting in terms of \(z\), the quadratic term will vanish.

Exercise 13

Find a solution to the cubic equation \(x^3 - 6x^2 + 27 x - 58 = 0\).

Historical note. A general method of solving cubic equations dates back to the work of Tartaglia and Cardano in the early 16th century. It's quite interesting that this work was done a century before Descartes introduced coordinate geometry!

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