Find the sum of all possible $a^3$, where $a$ is a rational figure.

Given that $a$ is rational and the equation $ax^2+(a+2)x+a-1=0$ has integer roots.

Find the sum of all possible $a^3$.

The solution will require deep and strategic thought but that doesn't mean this problem is impossible to solve.

If you know and are familiar with the manipulation trick that we could play on the rational number, we can see the solution pretty clearly.

Prove that $ab\leq \dfrac {1}{8}$ (Second Solution)

Given $b^2-4ac$ is a real root of equation :$ax^2+bx+c=0,\,\, (a\neq 0)$. Prove that $ab\leq \dfrac {1}{8}$.

Second solution is my solution: (I want to mention it here that the first solution is provided by a Taiwan friend of mine)

Multiply the equation of $ax^2+bx+c=0$ by $4a$, we get:

Prove that $ab\leq \dfrac {1}{8}$ (First Solution)

Given $b^2-4ac$ is a real root of equation :$ax^2+bx+c=0,\,\, (a\neq 0)$. Prove that $ab\leq \dfrac {1}{8}$.

Some students would find that this problem very confusing, as they have been told countless time that $b^2-4ac$  is actually a discriminant for the quadratic equation $ax^2+bx+c=0$.

In this instance, $b^2-4ac$ is both the discriminant and the root, they would think instantly something that looks like the following: