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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?

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May I ask where this cubic equation came from? – gorilla Jul 12 '11 at 7:22
up vote 2 down vote accepted

Yes, it's a bit tough to derive by hand; here's my Mathematica session on how I arrived at your expression:

{w,v,u}=FullSimplify[(Most[#]/Last[#])&[CoefficientList[(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, x]]];
q = FullSimplify[(u^2 - 3v)/9]; r = FullSimplify[(2u^3 - 9u v + 27w)/54];
y = -PowerExpand[(FullSimplify[r + PowerExpand[Sqrt[FullSimplify[r^2 - q^3]]]])^(1/3)];
FullSimplify[y + q/y -u/3]

(a - b)^2/(a^(2/3)*c^(1/3) + b^(2/3)*d^(1/3))^3
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Knowing a root in this case does help. Putting $x=y(a-b)^2$ and introducing new coefficients $A=(a^2 c)^{1/3}$, $B=(b^2 d)^{1/3}$ simplifies the equation.

Edit: For $z=y/(A+B)\ $ the equation becomes $z^3-3Cz^2+3z-1=0\ $ where $C=\frac{ A^2-7 A B+B^2}{(A+B)^2}$.

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I think you mean $x=y(a-b)^2$. – Gerry Myerson Jul 12 '11 at 7:02
Whoah, I wrote the same comment, character for character. Hivemind. – anon Jul 12 '11 at 7:04
You may mean $A=(a^2c)^{1/3}$ rather than $A=(ac^2)^{1/3}$ – Henry Jul 12 '11 at 7:05
@Gerry Yes, thanks. – Andrew Jul 12 '11 at 7:08

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