# Roots of a Certain type of Cubic Equation

Consider the following cubic equation in $x$ where all constants are positive. \begin{align} (x-a)(x-b)(x-c)-\alpha (x-a)-\beta (x-b) -abc=0 \end{align} Can any one show how to solve this cubic equation? I read the wikipedia article about solving cubic equations, but was wondering if there was a simple way to solve this particular as it had already some factorized form in it.

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I think the answer must be a disappointing "no". Suppose you set $b=c=0$ then the given form of cubic becomes: $$x^3-ax^2-(\alpha+\beta)x+\alpha a=0$$
Comparing this with the general cubic (monic form) $$x^3+px^2+qx+r=0$$
We find that we can put $a=-p, \alpha = -\frac rp, \beta = q+\frac rp$ provided we have $p\neq 0$ and transform the general cubic with $p\neq 0$ into a special version of the given form with $b=c=0$. If p=0, we can rewrite the equation by setting $y=x+1$, and write the equation in $y$ in the given form.