# Simplifying $\sqrt[3]{a\pm\sqrt{b}}$

Let $$x=\sqrt{a\pm\sqrt{b}}$$ We know that $$x=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$ But, what about cubic root?
Let $$y=\sqrt[3]{a\pm\sqrt{b}}$$ Is there any formula to find $c$ and $d$ such that $c,d\in\mathbb{Q}$ and $c\pm\sqrt{d}=y$ if $c$ and $d$ exists?
For example, let $$a=\sqrt[3]{45+\sqrt{1682}}$$ It can be solved factoring terms: $$a=\sqrt[3]{45+29\sqrt{2}}=\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}=\sqrt[3]{(3+\sqrt{2})^3}=3+\sqrt{2}$$ Is there any formula for cubic root like square root?

Simplifying $$\sqrt[3]{a+\sqrt{b}}$$. (WLOG. $$a>0$$.)

$$(n+m\sqrt{b})^3=(n^3+3nm^2b)+(3n^2m+m^3)\sqrt{b}.$$

$$n^3+3nm^2b=a, 3n^2m+m^3=1.$$

$$n=\sqrt{m^2+\dfrac{1}{3m}}.$$

$$a=\left(m^2+\dfrac 1 {3m} \right)^{\frac 1 2}\left( m^2(b+1)+\dfrac 1 {3m} \right)$$

$$a^2=\left( m^2+ \dfrac 1 {3m} \right)\left( m^2(b+1)+\dfrac 1 {3m} \right)^2.$$

$$27a^2m^3=(3m^3+1)(3m^3(b+1)+1)^2 \\ = 27(b+1)^2\left(m^3\right)^3+9(b+1)(b+1+2)\left(m^3\right)^2+3(2b+3)\left(m^3\right)+1.$$

$$\Rightarrow m^3=x, 27(b+1)^2x^3+9(b^2+4b+3)x^2+3(2b+3-a^2)x+1=0.$$

Solving this 3-dimensional equation,

$$\small x=\dfrac{1}{b+1}\left(\dfrac{a\sqrt{4b^3-a^2b^2+18a^2b-4a^4+27a^2}}{23^{\frac 9 2}}-\dfrac{3(b+1)+(a^2-2b-3)(b+3)}{162} - \dfrac{(b+3)^3}{3^6(b+1)^3}\right)^{\frac 1 3}+\dfrac{(b+3)^2-3(a^2-2b-3)}{9(b+1)\left(\dfrac{a\sqrt{4b^3-a^2b^2+18a^2b-4a^4+27a^2}}{23^{\frac 3 2}}-\dfrac{(b+3)^3}{27} + \dfrac{-3(b+1)-(a^2-2b-3)(b+3)}{6}\right)^{\frac 1 3}}-\dfrac{b+3}{3(b+1)}$$

...Which doesn't seem clear.

Anyway, $$m=x^{\frac 1 3}$$

$$=\left(\left({{a\,\sqrt{4\,b^3-a^2\,b^2+18\,a^2\,b-4\,a^4+27\,a^2} }\over{2\,3^{{{9}\over{2}}}\,\left(b+1\right)^3}}+{{-{{3}\over{27\,b ^2+54\,b+27}}-{{\left(a^2-2\,b-3\right)\,\left(b+3\right)}\over{ \left(3\,b+3\right)\,\left(9\,b^2+18\,b+9\right)}}}\over{6}}+{{ \left(-1\right)\,\left(b+3\right)^3}\over{27\,\left(3\,b+3\right)^3 }}\right)^{{{1}\over{3}}}-{{{{\left(-1\right)\,\left(b+3\right)^2 }\over{9\,\left(3\,b+3\right)^2}}-{{a^2-2\,b-3}\over{3\,\left(9\,b^2 +18\,b+9\right)}}}\over{\left({{a\,\sqrt{4\,b^3-a^2\,b^2+18\,a^2\,b- 4\,a^4+27\,a^2}}\over{2\,3^{{{9}\over{2}}}\,\left(b+1\right)^3}}+{{- {{3}\over{27\,b^2+54\,b+27}}-{{\left(a^2-2\,b-3\right)\,\left(b+3 \right)}\over{\left(3\,b+3\right)\,\left(9\,b^2+18\,b+9\right)}} }\over{6}}+{{\left(-1\right)\,\left(b+3\right)^3}\over{27\,\left(3\, b+3\right)^3}}\right)^{{{1}\over{3}}}}}+{{\left(-1\right)\,\left(b+3 \right)}\over{3\,\left(3\,b+3\right)}}\right)^{\frac 1 3}$$

and $$n=\sqrt{m^2+\dfrac 1 m}$$

$$= \left(\left({{a\,\sqrt{4\,b^3-a^2\,b^2+18\,a^2\,b-4\,a^4+27\,a^2} }\over{2\,3^{{{9}\over{2}}}\,\left(b+1\right)^3}}+{{-{{3}\over{27\,b ^2+54\,b+27}}-{{\left(a^2-2\,b-3\right)\,\left(b+3\right)}\over{ \left(3\,b+3\right)\,\left(9\,b^2+18\,b+9\right)}}}\over{6}}+{{ \left(-1\right)\,\left(b+3\right)^3}\over{27\,\left(3\,b+3\right)^3 }}\right)^{{{1}\over{3}}}-{{{{\left(-1\right)\,\left(b+3\right)^2 }\over{9\,\left(3\,b+3\right)^2}}-{{a^2-2\,b-3}\over{3\,\left(9\,b^2 +18\,b+9\right)}}}\over{\left({{a\,\sqrt{4\,b^3-a^2\,b^2+18\,a^2\,b- 4\,a^4+27\,a^2}}\over{2\,3^{{{9}\over{2}}}\,\left(b+1\right)^3}}+{{- {{3}\over{27\,b^2+54\,b+27}}-{{\left(a^2-2\,b-3\right)\,\left(b+3 \right)}\over{\left(3\,b+3\right)\,\left(9\,b^2+18\,b+9\right)}} }\over{6}}+{{\left(-1\right)\,\left(b+3\right)^3}\over{27\,\left(3\, b+3\right)^3}}\right)^{{{1}\over{3}}}}}+{{\left(-1\right)\,\left(b+3 \right)}\over{3\,\left(3\,b+3\right)}}\right)^{\frac 2 3}+\left(\left({{a\,\sqrt{4\,b^3-a^2\,b^2+18\,a^2\,b-4\,a^4+27\,a^2} }\over{2\,3^{{{9}\over{2}}}\,\left(b+1\right)^3}}+{{-{{3}\over{27\,b ^2+54\,b+27}}-{{\left(a^2-2\,b-3\right)\,\left(b+3\right)}\over{ \left(3\,b+3\right)\,\left(9\,b^2+18\,b+9\right)}}}\over{6}}+{{ \left(-1\right)\,\left(b+3\right)^3}\over{27\,\left(3\,b+3\right)^3 }}\right)^{{{1}\over{3}}}-{{{{\left(-1\right)\,\left(b+3\right)^2 }\over{9\,\left(3\,b+3\right)^2}}-{{a^2-2\,b-3}\over{3\,\left(9\,b^2 +18\,b+9\right)}}}\over{\left({{a\,\sqrt{4\,b^3-a^2\,b^2+18\,a^2\,b- 4\,a^4+27\,a^2}}\over{2\,3^{{{9}\over{2}}}\,\left(b+1\right)^3}}+{{- {{3}\over{27\,b^2+54\,b+27}}-{{\left(a^2-2\,b-3\right)\,\left(b+3 \right)}\over{\left(3\,b+3\right)\,\left(9\,b^2+18\,b+9\right)}} }\over{6}}+{{\left(-1\right)\,\left(b+3\right)^3}\over{27\,\left(3\, b+3\right)^3}}\right)^{{{1}\over{3}}}}}+{{\left(-1\right)\,\left(b+3 \right)}\over{3\,\left(3\,b+3\right)}}\right)^{-\frac 1 3}$$

Last comment: Maybe it should be better to try some possible cases, rather than formulizing it.

• I think your first expansion is wrong, you're lacking the 3s: $(n+m\sqrt b)^3 = n^3 + m^3b\sqrt b + 3nm\sqrt b (n + m \sqrt b)$ Commented Jun 3, 2023 at 15:30
• Alright, then I'll fix $\sqrt[3]{a+\sqrt{b}}$ to $\sqrt[3]{a+3\sqrt{b}}$ to minimize my edits... Sorry for that.
– RDK
Commented Jun 3, 2023 at 15:49
• I'm afraid the 3 are in the wrong place still, one of them have the $\sqrt b$ and the other doesn't. Sorry. Commented Jun 3, 2023 at 15:56
• @hellofriends Oh, I'll fix it, so that my answer can be perfect!
– RDK
Commented Jun 4, 2023 at 16:29
• I'm so very sorry, but the $m^3$ inside of the first parenthesis should be $m^3b$. Commented Jun 4, 2023 at 16:44

There isn't a formula for $\sqrt[3]{a\pm\sqrt{b}}$ The best method that I know is simplifying $\sqrt{b}$ (if possible) and assuming that it can be denested into $x+y\sqrt{b}$.

More generally, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}$$

$$y$$ is a real root of a 6th degree polynomial with rational coefficients:

$$y=\sqrt[3]{a\pm\sqrt{b}}$$ $$y^3-a=\pm\sqrt{b}$$ $$y^6-2ay^3+\left(a^2-b\right)=0$$

You would like $$y$$ to equal $$c+\sqrt{d}$$, a real root of a quadratic polynomial: $$y=c+\sqrt{d}$$ $$y-c=\sqrt{d}$$ $$y^2-2cy+(c^2-d)=0$$

So it depends on whether the rational factorization of $$y^6-2ay^3+\left(a^2-b\right)$$ has a quadratic factor. (If it does, then you can directly solve for $$c$$ using the linear term, and then solve for $$d$$ using the constant term.)

Degree 6 polynomials do not, in general, have a quadratic factor. But what about this one? It factors over $$\mathbb{R}$$ as two cubics:

$$\left(y^3-a-\sqrt{b}\right)\left(y^3-a+\sqrt{b}\right)$$

And these cubics factor with one real root and two rotations:

$$\left(y-\sqrt[3]{a+\sqrt{b}}\right)\left(y-\omega\sqrt[3]{a+\sqrt{b}}\right)\left(y-\omega^2\sqrt[3]{a+\sqrt{b}}\right)\left(y-\sqrt[3]{a-\sqrt{b}}\right)\left(y-\omega\sqrt[3]{a-\sqrt{b}}\right)\left(y-\omega^2\sqrt[3]{a-\sqrt{b}}\right)$$

You would like some pair of these factors that includes a real root (so includes either the 1st or 4th factor) to multiply together to make a quadratic with rational coefficients. That is only possible using precisely the 1st and 4th factor.

$$\left(y-\sqrt[3]{a+\sqrt{b}}\right)\left(y-\sqrt[3]{a-\sqrt{b}}\right)=y^2-\left(\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}\right)y+\sqrt[3]{a^2-b}$$

So you have to be in the lucky position that $$a^2-b$$ has a rational cube root. And furthermore, that $$\left(\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}\right)$$ is rational. It's unlikely with random $$a,b$$, but possible every now and then like with $$a=7,b=50$$.

Your question asks "if $$c$$ and $$d$$ exist". So if the two conditions are met in the last paragraph, then yes. You would have:

\begin{align} c&=\frac12\left(\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}\right)\\ d&=c^2-\sqrt[3]{a^2-b} \end{align}

That is:

$$\sqrt[3]{a+\sqrt{b}}=\frac12\left(\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}\right)+\sqrt{\frac14\left(\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}\right)^2-\sqrt[3]{a^2-b}}$$

• "And furthermore, that $\left(\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}\right)$ is rational." $\;-\;$ FWIW a sufficient condition for that to happen is given in my answer here: "a sufficient condition for $\,a,b = \sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$ to denest is for $\,m^2 \cdot p - n^2 \cdot q\,$ to be the cube of a rational $\,r\,$, and for the cubic $\,p\, t'^{\,3} - 3r\, t' - 2m\,$ to have an appropriate rational root, and in that case $\,a,b = \frac{1}{2}\left(t'\,\sqrt{p} \pm \sqrt{t'^{\,2} p-4r}\right)\,$".
– dxiv
Commented Jun 3, 2023 at 23:22