<s>[INST] 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
 [/INST] 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.
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