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
First Answer:
\begin{align*}
2\alpha +\beta = -\frac{3b}{a} &amp;\Rightarrow \beta=-\frac{3b}{a}-2\alpha\\
&amp;\Rightarrow {\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\\
&amp;\Rightarrow -3\alpha^2-\frac{6b}{a}\alpha=\frac{3c}{a} \\
&amp;\Rightarrow a{\alpha}^2+2b\alpha+c=0\tag{0}\\
&amp;\Rightarrow 2a{\alpha}^3+4b{\alpha}^2+2c\alpha=0\tag{1}
\end{align*}
and 
\begin{align*}
{\alpha}^2\left(-2\alpha-\frac{3b}{a}\right)=-\frac da\Rightarrow 2a{\alpha}^3+3b{\alpha}^2-d=0.\tag{2}
\end{align*}

By subtracting (1) and (2) we have
 $$b\alpha^2+2c\alpha+d=0\Rightarrow ab\alpha^2+2ac\alpha+ad=0.$$
Now from (0) we have 
$$ab\alpha^2+2b^2\alpha+bc=0$$
Now subtract two last equations: $(2ac-2b^2)\alpha+ad-bc=0$
Thus
$$\fbox{$\alpha=\frac{bc-ad}{2(ac-b^2)}$}$$

Second Answer for alpha:

You can continue like this
$$
2\alpha +\beta = -\frac{3b}{a} \Rightarrow \beta=-\frac{3b}{a}-2\alpha\\
$$
put it in the last one: 
\begin{align*}
\left(-\frac{3b}{a}-2\alpha\right)\alpha^2=-\frac{d}{a}&amp;\Rightarrow-2\alpha^3-\frac{3b}{a}\alpha^2=-\frac{d}{a}\\
&amp;\Rightarrow 2a\alpha^3+3b\alpha^2-d=0\\
\end{align*}
on the other hand $\alpha$ must satisfy the original equation, thus
$$a\alpha^3+3b\alpha^2+3c\alpha+d=0$$
Then add two last equations:
$$3a\alpha^3+6b\alpha^2+3c\alpha=0$$
Since $\alpha\ne0$ we have : $a\alpha^2+2b\alpha+c=0$, thus
$$\alpha=-b+\sqrt{b^2-ac}\quad \text{ or }\quad \alpha=-b-\sqrt{b^2-ac} $$
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 $\frac{-b+\sqrt[3]{b^3 - 27a^2d}}{3a}$.