The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation. The given question is:
The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation.
I am confused about how to work out this question, but I feel like it has something to do with using α,β,γ, δ  as roots of the quartic.
I would greatly appreciate all help in a simple and efficient method to solving this question. Please note, according to the answers, there is only one value of $a$ and one value of $b$.

We know that the equation has a triple root, so if let $\alpha$ be the triple root and $\beta$ be the single root, we get that
\begin{equation}
x^4 - 3x^3-6x^2 + ax + b = (x-\alpha)^3(x-\beta)
\end{equation}
By expanding the RHS, we see that
\begin{equation}
x^4 - 3x^3-6x^2 + ax + b = x^4 - (3\alpha+\beta)x^3 + (3\alpha\beta + 3\alpha^2)x^2 + \cdots 
\end{equation}
Now equating the coefficients of the $x^2$ and $x^3$ term, we have the simultaneous equations
\begin{equation}
3 = 3\alpha+\beta \quad ; \quad -6 = 3\alpha\beta + 3\alpha^2
\end{equation}
By a simple substitution we conclude $(\alpha,\beta) = (2,-3)$ or $(\alpha,\beta) = (-1/2, 9/2)$. We can check that these indeed satisfy our conditions. It then follows that we have either $(a,b) = (28,-24)$ or  $(a,b) = (-13/4,-9/16)$.
