If $a @ b=\frac{a^{3}-b^{3}}{a-b}$, for how many real values of $a$ does $a @ 1=0$ ?
Solution: If $\frac{a^{3}-1}{a-1}=0$, then $a^{3}-1=0$, or $(a-1)\left(a^{2}+a+1\right)=0$. 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 $\mathbf{0}$.