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Given that $x^2 + x + 1 = 0$, find the value of $x^3$.


Begin by multiplying through by $x$:

$$x^3 + x^2 + x = 0$$

Adding one to both sides:

$$x^3 + x^2 + x + 1 = 1$$

But as $x^2 + x + 1 = 0$ we deduce that $x^3 = 1$.

What is interesting about this problem is that knowing too much could make it more difficult to solve. If you try to solve the original equation you will encounter the square root of a negative number. For most students this is the end of the road. But eventually students learn how to solve these and would obtain the complex roots: $x = \frac{1}{2}(-1 \pm \sqrt{3} i)$. But even then cubing these roots is rather tedious and would only lead to the result we established in a couple of simple steps.

However, it must be noted that we cannot simply use the result we obtained above: $x^3 = 1$, to claim that $x = \sqrt[3]{1} = 1$ is a root of the original equation; try it out and you'll see that it doesn't "work". It turns out (using Advanced Mathematics) that the cube root of unity has three values: the complex roots given above and 1. But in this case it is only the two complex roots that satisfy the original equation.

The reason this happens is related to a common trick used in proving $1 = 0$ or similar absurd results. The secret is to watch out for the point in the "proof" where the extra root (which doesn't belong to the original equation) is introduced. If the original equation is a quadratic then it has two roots. If as part of our algebraic manipulation we produce a cubic then that cubic may have up to three distinct roots of which only two can belong to the original equation.

See if you can spot the flaw in the following related fallacious proof.

Suppose that $x^2 + x + 1 = 0$.

We can rearrange differently to get $x + 1 = -x^2$ and $x(x + 1) = -1$.

Substituting $-x^2$ for $x + 1$ in the second equation we get $-x^3 = -1 \Rightarrow x = 1$.

But try substituting $x = 1$ into the original equation... it seems that $3 = 0$?!

Problem ID: 354 (23 Jul 2009)     Difficulty: 3 Star

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