Sad Cubic Equation

This is better solved graphically. The solutions to the equation are the intersections between the curve $(C): y = \displaystyle\frac {x^3} {3} - x$ and the line $(d): y=b$ .

As $(d)$ is parallel to the horizontal axis and $(C)$ is a N-shaped curve, there are three intersections (hence three distinct solutions) if and only if $(d)$ lies between the two extrema, excluding the extrema themselves.

As $y'= x^2 - 1 = 0 \Leftrightarrow x = \pm 1$ , the extrema are $\left(-1,\displaystyle\frac {2} {3}\right)$ and $\left(1,-\displaystyle\frac {2} {3}\right)$ . Therefore, $-\displaystyle\frac {2} {3} < b < \displaystyle\frac {2} {3}$ .

The only intergral value of $b$ that satisfies this is $b = 0$ . Hence, only $\boxed {1}$ value of $b$ satisfies the solution.

A monic cubic polynomial without quadratic term $x^3 + px + q$ has discriminant $Δ = -4p^3 - 27q^2$ .

For roots to be distinct, Δ should be greater than $0$ .

$\implies (-4(-3^{3}) )-27(9b^2) > 0$

$\implies \frac{2}{3} > b > \frac{-2}{3}$ .

$\implies$ The only integral value of $b$ is $0$

Hence total number of integral values = $1$ .

Using Graphs......