Solving $d \cdot a^4 - 2 \cdot d^2 \cdot a^2 + d^3 - a^2 = 0$ for $d$

Question

Can anybody help me with this equation? I can't find a way to factorize for finding a value of $d$ as a function of $a$:

$$d^3 - 2\cdot d^2\cdot a^2 + d\cdot a^4 - a^2 = 0$$

Another form:

$$d=\frac{a^2}{(a^2-d)^2}$$

Maybe this equation has no solution. I don't know. That equation is out of some calculus involving the golden number.

Thx for your help.

Answer 1

If you're really keen on solving for $d$ in terms of $a$ and getting an explicit answer, there's no help for it except solving a cubic. I'm going to help take you through how to solve a cubic equation in general. Note that you'll have to plug in the actual values and work through the detailed algebra yourself.

So first note that your equation is a monic cubic in $d$. Monic means the lead coefficient is one. It's easy to make any cubic a monic by division, but since this is already in that form, it makes life a little easier.

Let's say you have a monic cubic $x^3 + ax^2 + bx + c = 0$ (sorry about the re-use of $a$ here, but this is the form that I have handily written down when I worked out the solution and kept it in my google drive).

First step is to substitute $x = y-\frac a3$

That will allow you to reduce the equation to this form:

$y^3 + \alpha y + \beta = 0$.

where you can express $\alpha$ and $\beta$ in terms of the original coefficients. ($\alpha = b - \frac {a^2}{3}, \beta = \frac{2a^3}{27} - \frac{ab}{3}+c$, if you must know).

This form is called a depressed cubic as it lacks the square term. It's much easier to solve.

You now have to make the substitution $y = z - \frac{\alpha}{3z}$

That will reduce the equation to $z^3 - \frac k{z^3} + \beta = 0$.

(where $k$ is expressible in terms of $\alpha$. Again, if you must know, $k = \frac{\alpha^3}{27}$).

Finally substitute $m = z^3$.

This allows you to get to a quadratic:

$m^2 + \beta m - k = 0$

which can be solved exactly as $m = \frac{-\beta \pm \sqrt{\beta^2 + 4k}}2$

And of course, you can now work backwards: from $m$ to $z$ to $y$ to $x$.

As I said, very tedious, but doable.

The alternative, as mentioned in the comments is to express $a^2$ in terms of $d$ using the quadratic formula, but this may not be what you're looking for.

Answer 2

The equation can be rewritten as $$d^3-2a^2d^2+a^4d-a^2=0,$$ which is a cubic in $d$. This can be solved by applying Cardano's formula to the depressed cubic $$d^3-2a^2d^2+a^4d-a^2=x^3-a^4x+a^2\left(\tfrac{26}{27}a^4-1\right),$$ where $x:=d-\tfrac23a^2$. Then $$x=\frac{a^{2/3}}{3\sqrt[3]{2}}\sqrt[3]{27-26a^4+\sqrt{568a^8-1404a^4+729}}+\frac{a^{2/3}}{3\sqrt[3]{2}}\sqrt[3]{27-26a^4-\sqrt{568a^8-1404a^4+729}},$$ where the cube roots are of real numbers if and only if $$568a^8-1404a^4+729\geq0,$$ or equivalently $\left|a^4-\frac{351}{284}\right|\geq\frac{81}{284}\sqrt{3}$. Then $d=\frac23a^2+x$.