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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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up vote 3 down vote accepted

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.

If the given form were easy to factorise, we could therefore transform any cubic into that form quite straightforwardly, and then solve it. But solving the cubic is not that easy (and in general will involve taking both square and cube roots).

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