Cubic equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots. How can I find their value in terms of $a,b,c$? If the equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots, then equal root must
be equal to $\displaystyle \frac{bc-ad}{2(ac-b)^2}.$
My Try:: Let $x=\alpha,\alpha,\beta$ be the roots of given equation. Then using Vieta's formula
$$ \alpha+\alpha+\beta = -\frac{3b}{a}\Rightarrow 2\alpha +\beta = -\frac{3b}{a}$$
$$ \alpha \cdot \alpha +\alpha \cdot \beta +\alpha \cdot \beta = \frac{3c}{a}\Rightarrow \alpha^2+2\alpha \cdot \beta = \frac{3c}{a}$$
$$\alpha \cdot \alpha \cdot \beta = -\frac{d}{a}\Rightarrow \alpha^2 \cdot \beta = -\frac{d}{a}.$$
Now I did not understand how can I find the value of $\alpha$ in terms of $a,b$ and $c$.
Help is required.
Thanks

I made a formula to solve a cubic equation with all equal roots.
$$ax^3+bx^2+cx+d=0$$
To find such an equation you will have to check the condition $b=\sqrt{3ac}$ if it is so then we will use this formula $\frac{-b+\sqrt[3]{b^3 - 27a^2d}}{3a}$.
