If $a@b=\dfrac{a^3-b^3}{a-b}$, for how many real values of $a$ does $a@1=0$?
I notice that the expression $a^3-b^3$ can be factored as $(a-b)(a^2+ab+b^2)$. So, $a@b=\dfrac{a^3-b^3}{a-b}=\dfrac{(a-b)(a^2+ab+b^2)}{a-b}$. If $a\neq b$, I can cancel out the $(a-b)$ factors and get $a@b=a^2+ab+b^2$. But if $a=b$, then $a@b$ is undefined, because I would be dividing by zero. So, I need to be careful not to let $a=b$ when I solve for $a$. Now, I want to find $a$ such that $a@1=0$. That means $a^2+a+1^2=0$. This is a quadratic equation in $a$. To solve it, I can use the quadratic formula: $a=\dfrac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}$. Simplifying, I get $a=\dfrac{-1\pm\sqrt{-3}}{2}$. The square root of a negative number is not real, so $a$ is not real either. There are no real values of $a$ that make $a@1=0$. 