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
 A: First Answer:
\begin{align*}
2\alpha +\beta = -\frac{3b}{a} &\Rightarrow \beta=-\frac{3b}{a}-2\alpha\\
&\Rightarrow {\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\\
&\Rightarrow -3\alpha^2-\frac{6b}{a}\alpha=\frac{3c}{a} \\
&\Rightarrow a{\alpha}^2+2b\alpha+c=0\tag{0}\\
&\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}&\Rightarrow-2\alpha^3-\frac{3b}{a}\alpha^2=-\frac{d}{a}\\
&\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
$$\alpha=-b+\sqrt{b^2-ac}\quad \text{ or }\quad \alpha=-b-\sqrt{b^2-ac} $$
A: From 1 , take $\beta = -\frac{3b}a - 2\alpha$
Substitute $\beta$ in 3, 
$$\alpha^2 \left(-\frac{3b}a - 2\alpha\right) = -\frac da$$
You will get a cubic in $\alpha$ which you can solve.
Or you can substitute the value of $\beta$ in 2 from where you will get a  quadratic in $\alpha$ which will be easier to solve.
A: Since $\beta=-2\alpha-\frac{3b}{a}$, we have 
$${\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\Rightarrow a{\alpha}^2+6b\alpha+c=0\Rightarrow 2a{\alpha}^3+12b{\alpha}^2+2c\alpha=0,$$
$${\alpha}^2\left(-2\alpha-\frac{3b}{a}\right)=-\frac da\Rightarrow 2a{\alpha}^3+3b{\alpha}^2-d=0.$$
Hence, we have
$$9{\alpha}^2+2c\alpha+d=0\Rightarrow 9a{\alpha}^2+2ca\alpha+ad=0.$$
Hence, we've already have
$$9a{\alpha}^2+54b\alpha+9c=0,$$
we have
$$(2ca-54b)\alpha=9c-ad$$
Hence, we have
$$\alpha=\frac{9c-ad}{2(ca-27b)}.$$
I don't know how to reach your value.
A: I made another formula to find roots of cubic eq with two equal and one unequal roots.
x^3+bx^2+cx+d
The formula is
Root 1(two equal roots)= - { b (+/-)√(b^2 - 3c)/ 3}
Root 2 (unequal) =  - { -b (+/-)2√(b^2 - 3c)/ -3}
However, you yourself have to identify such cubic eq
whether it has two equal roots.you will be confused by +/- but one of the operators will give correct ans while one has to be neglected. To find which ans is correct just put the value and check which one is the zero of the eq.
A: 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 $\frac{-b+\sqrt[3]{b^3 - 27a^2d}}{3a}$.
