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.
On
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:
$$\begin{cases}I&\alpha^3-b\alpha+a=0\\{}\\II&3\alpha^2-b=0\end{cases}\;\stackrel{I I}\implies \alpha=\sqrt\frac b3\stackrel I\implies$$
$$\stackrel I\implies \left(\frac b3\right)^{3/2}-\frac{b^{3/2}}{\sqrt3}+a=0\implies$$
$$a=\frac{-b\sqrt b+3b\sqrt b}{3\sqrt3}=2\left(\frac b3\right)^{3/2}$$
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.