I have this not so great looking cubic polynomial in $x$:
$(a - b)^6 + 3 (a - b)^4 (-a^2 c - b^2 d) x + 3 (a - b)^2 (a^4 c^2 - 7 a^2 b^2 c d + b^4 d^2) x^2 + (-a^2 c - b^2 d)^3 x^3$
where $a,b,c,d>0$ and $c+d=1$ and I need to find a real root of the above cubic (in fact there should be only one, as I know that the discriminant is negative). I know there are general solutions and Mathematica can easily whip out a complicated solution for it (which it does).
Now, I know that the following nifty expression is a root (the root I'm looking for):
$x_1 = \frac{(a - b)^2}{\left((a^2 c)^{1/3} + (b^2 d)^{1/3}\right)^3}$
and I see patterns of the solution in the coefficients of my cubic. However, I can't seem to obtain $x_1$ from my cubic in that form.
Here's the cubic and the solution in Mathematica code:
eqn = (a - b)^6 + 3 (a - b)^4 (-a^2 c - b^2 d) x + 3 (a - b)^2 (a^4 c^2 - 7 a^2 b^2 c d + b^4 d^2) x^2 + (-a^2 c - b^2 d)^3 x^3; sol = ((a - b)^2)/((a^2 c)^(1/3) + (b^2 d)^(1/3))^3; (*verify sol is a root of eqn*) In[3]:= Simplify[eqn /. x -> sol] Out[3]= 0
As to how I know that it is a root, I found it in a paper (non-math) with no explanation other than "tedious algebra" and mailing the authors didn't help. I'll agree it was tedious to work out the math and arrive at this cubic, but I feel like I'm almost at the finish line and need a little push to cross the line.
In my end use application, it's going to be a numerical routine and for my needs, it doesn't matter if it's one expression or the other as long as they are accurate to a certain tolerance. However, there's a certain elegance to the above solution and I'm very interested in nailing it down. I tried using the general formula for the roots of a cubic but it doesn't seem like a pretty place to begin. Are there alternate ways to approach this?