Problem 628

Solution:

If $\frac{a^{3} - 1}{a - 1} = 0$, then $a^{3} - 1 = 0$, or $(a - 1)(a^{2} + a + 1) = 0$.

Thus $a = 1$, which is an extraneous solution since that makes the denominator of the original expression $0$, or $a$ is a root of $a^{2} + a + 1$.

But this quadratic has no real roots, in particular its roots are $\frac{-1 \pm \sqrt{-3}}{2}$.

Therefore there are no such real values of $a$, so the answer is $\mathbf{0}$.