Solution for cubic algebra For a cubic equation:
$$x^3-xb+a=0 \\$$
EDIT: the above equation has three real solutions for x.
one of the solutions is:
$$a=2\cdot \left(\frac b3\right)^{3/2}$$
EDIT: "one of the solutions is:" should read, "by solving for x, the answer to "a" was found. 
How does one arrive at this?
Could anyone please show me the working out please.
 A: I think I understood what you meant. What you wrote is certainly not a solution of the cubic. It is a condition. For the equation $$x^3 - bx + a = 0$$ to have only real solutions, you need the determinant to be positive, that is : $$\frac{a^2}{4} - \frac{b^3}{27}\ge 0$$
It seems that you want to solve the equality $\dfrac{a^2}{4} - \dfrac{b^3}{27} = 0$ which is solved when $a = 2\bigg(\dfrac{b}{3}\bigg)^{3/2}$ which is what you have.
When the determinant is zero, or when the above equation is satisfied, then you have a exactly two (distinct) real solutions as your expression can be factorised : $$x^3 - bx + 2\bigg(\dfrac{b}{3}\bigg)^{3/2} = \bigg(x - \sqrt{\frac{b}{3}}\bigg)^2\bigg(x + 2\sqrt{\frac{b}{3}}\bigg)$$
Here's an illustrative plot using WA which shows what happens when $b = 5$ for example.
A: As one of the roots is multiple (why? Otherwise we cannot deduce an equality for $\;a\;$ by only looking at the cubic's discriminant), the cubic's derivative also vanishes at the multiple root  $\;\alpha\;$ , say:
$$\begin{cases}I&\alpha^3-b\alpha+a=0\\{}\\II&3\alpha^2-b=0\end{cases}\;\stackrel{I I}\implies \alpha=\sqrt\frac b3\stackrel I\implies$$
$$\stackrel I\implies \left(\frac b3\right)^{3/2}-\frac{b^{3/2}}{\sqrt3}+a=0\implies$$
$$a=\frac{-b\sqrt b+3b\sqrt b}{3\sqrt3}=2\left(\frac b3\right)^{3/2}$$
