# The Short Version

Is there a way to simplify this expression?

$$\left(\left(\left(d × \left(\frac{j}{2}\right)^2\right)^2 − \frac{1}{27} × \left(\frac{j^2}{2 s}\right)^6\right)^\frac{1}{2} + d × \left(\frac{j}{2}\right)^2\right)^\frac{1}{3} + \frac{1}{3} × \left(\frac{j^{2}}{2 s}\right)^2 × \left(\left(\left(d × \left(\frac{j}{2}\right)^2\right)^2 − \frac{1}{27} × \left(\frac{j^2}{2 s}\right)^6\right)^\frac{1}{2} + d × \left(\frac{j}{2}\right)^2\right)^\frac{−1}{3} − \frac{j^2}{2 s}$$

Specifically, I'd like to condense the cube root portion and the inverse cube root portion into a single root, if possible.

# The Longer Version

We have five positive-valued variables:

• $$s_{max}$$
• $$j_{max}$$
• $$a_{max}$$
• $$v_{max}$$
• $$d_{max}$$

We also have the function $$\operatorname{Min}(n_1,\, n_2,\, …,\, n_k)$$, which returns the input with the lowest value.

Finally, we have three variables whose values are each based on combinations of the previously-defined variables:

$$j_{limit}\,=\,{\operatorname{Min}\begin{pmatrix} \left(j_{max} × \frac{s_{max}^0}{0!}\right)^\frac{1}{1},\\ \left(a_{max} × \frac{s_{max}^1}{1!}\right)^\frac{1}{2},\\ \left(v_{max} × \frac{s_{max}^2}{2!}\right)^\frac{1}{3},\\ \left(d_{max} × \frac{s_{max}^3}{3!}\right)^\frac{1}{4} \end{pmatrix}}$$

$$a_{limit}\,=\,{\operatorname{Min}\begin{pmatrix} a_{max},\\ \left(v_{max} × j_{limit} + \left(\frac{j_{limit}^2}{2 s_{max}}\right)^2\right)^\frac{1}{2} - \frac{j_{limit}^2}{2 s_{max}},\\ \left(\left(\left(d_{max} × \left(\frac{j_{limit}}{2}\right)^2\right)^2 − \frac{1}{27} × \left(\frac{j_{limit}^2}{2 s_{max}}\right)^6\right)^\frac{1}{2} + d_{max} × \left(\frac{j_{limit}}{2}\right)^2\right)^\frac{1}{3} + \frac{1}{3} × \left(\frac{j_{limit}^{2}}{2 s_{max}}\right)^2 × \left(\left(\left(d_{max} × \left(\frac{j_{limit}}{2}\right)^2\right)^2 − \frac{1}{27} × \left(\frac{j_{limit}^2}{2 s_{max}}\right)^6\right)^\frac{1}{2} + d_{max} × \left(\frac{j_{limit}}{2}\right)^2\right)^\frac{−1}{3} − \frac{j_{limit}^2}{2 s_{max}} \end{pmatrix}}$$

and

$$v_{limit}\,=\,{\operatorname{Min}\begin{pmatrix} v_{max},\\ \left(d_{max} × a_{limit} + \left(\frac{a_{limit}^2}{2 j_{limit}}\right)^2\right)^\frac{1}{2} - \frac{a_{limit}^2}{2 j_{limit}} \end{pmatrix}}$$

The third option for the value of $$a_{limit}$$ stands out for being so much longer than all the other expressions in the values of the limit-variables. Can it be condensed any? Is there a better way to write it? As-is, it feels very unsatisfying.

• There are a large number of incredibly basic algebraic steps that could be made to your expression (expanding powers of monomials, factoring common factors, distributing roots over products, etc.); these will obviously make what you have look much simpler. Whether at the end of the day they will resolve the major ways in which your expression is complicated (the nested roots) is less clear. I'm personally not inclined to go through the trouble of the tedious arithmetic to get to the potentially interesting part.
– JBL
Apr 22, 2023 at 15:31
• TLDR. if you want assistance, try stating your question conscisely. Apr 24, 2023 at 20:58

Let $$\,\displaystyle a = d \cdot \left(\frac{j}{2}\right)^2\,$$ and $$\,\displaystyle b = \frac{1}{3} \cdot \left(\frac{j^2}{2s}\right)^2\,$$, then the expression (without the radical-free term at the end) is:

\require{cancel} \begin{align} x &= \sqrt[3]{a + \sqrt{a^2 − b^3}} + \frac{b}{\sqrt[3]{a + \sqrt{a^2 − b^3}}} \cdot \color{red}{\frac{\sqrt[3]{a - \sqrt{a^2 − b^3}}}{\sqrt[3]{a - \sqrt{a^2 − b^3}}}} \\ &= \sqrt[3]{a + \sqrt{a^2 − b^3}} + \frac{ \bcancel{b}\,\sqrt[3]{a - \sqrt{a^2 − b^3}}}{\bcancel{\sqrt[3]{\cancel{a^2} - (\cancel{a^2} − b^3)}}} \\ &= \sqrt[3]{a + \sqrt{a^2 − b^3}} + \sqrt[3]{a - \sqrt{a^2 − b^3}} \end{align}

Let $$\,u,v = \sqrt[3]{a \pm \sqrt{a^2 − b^3}}\,$$, then $$\,uv = b\,$$, and:

\begin{align} x^3 = (u+v)^3 &= u^3+v^3+3uv(u+v) \\ &= \left(a + \cancel{\sqrt{a^2 − b^3}}\right) + \left(a - \cancel{\sqrt{a^2 − b^3}}\right) + 3b(u+v) \\ &= 2a + 3 b x \end{align}

Therefore $$\,x\,$$ is a root of the depressed cubic $$\,x^3 - 3bx - 2a = 0\,$$, and the initial expression is in fact one form of Cardano's formula. Reverting to the original parameters, the equation is:

$$x^3 - \cancel{3} \cdot \frac{1}{\cancel{3}} \cdot \left(\frac{j^2}{2s}\right)^2 x - 2 \cdot d \cdot \left(\frac{j}{2}\right)^2 = 0 \quad\iff\quad 4 s^2 \,x^3 - j^4 \,x - 2 d j^2s^2 = 0$$

The latter is equivalent to the general depressed cubic, and there is no known formula simpler than the one that contains two cube roots (unless there are additional constraints on $$d,j,s$$).

• Congratulation on taking up this question. Apr 24, 2023 at 21:01
• @NoChance Thanks. Thought that would be fair, since the OP put a lot of effort in asking the question.
– dxiv
Apr 24, 2023 at 21:09
• @dxiv I can follow along with what you're doing in the first half, but I'm confused as to what's going on after you introduce $u$, $v$, and $x$. I understand each step individually - that is, the mathematical operations make sense - but I'm not sure how it connects to my question. What am I missing? Apr 25, 2023 at 14:56
• @dxiv Technically, there is a way to solve the general cubic using only a single cube root. (It uses a Tschirnhausen transformation.) I'll give a second answer to this question within the day or so to illustrate. Jun 20, 2023 at 16:58
• @TitoPiezasIII Thanks for raising the point. Guess I should have been more careful with wording the "it is not possible..." part. What I really meant was "it is not possible in a practical way...". You are technically right, and I imagine that an appropriately chosen quadratic transformation could do that. It will be interesting to see the final form, I'd expect a few more square roots around.
– dxiv
Jun 20, 2023 at 20:46