# Generalized denesting formula for $\sqrt[3]{A+B\sqrt{C}}$

Question: What is the generalized formula for denesting

$$\sqrt[3]{A+B\sqrt{C}}$$

Recently I've posted a question about nested radicals in solving the cubic equation. I received a comment about an ingenious method to denest the radical by solving a cubic equation, but, I was still not satisfied. See $\sqrt[3]{\text{something}\pm\sqrt{\text{something}}}$ and the link from the comments Denesting Phi, Denesting Cube Roots.

Trying to solve cubic equations, sometimes I would get something like $$\sqrt[3]{-27+6\sqrt{21}}$$, which is not easily denested by hand. I had to go and solve another cubic equation. The radical derives from the equation $$x^3+x-2=0$$. The cubic equation I had to solve for the radical would be more complex than the equation itself. I tried searching the internet for a denesting formula but I couldn't find a definitive one. All I could find was something about Galois theory which is mathematics I don't understand (I'm a 10th grader in Romania).

So, I tried making my own algorithm. Based on the idea from Denesting Phi, Denesting Cube Roots, I might have generalized it to solving a depressed cubic equation, which should be relatively easy and, in lucky cases, immediate. Keep in mind this is not a formula, but a simplification.

$$\sqrt[3]{A+B\sqrt{C}}$$ We want to write $$A+B\sqrt{C}$$ as $$(a+b\sqrt{C})^3$$. $$(a+b\sqrt{C})^3=(a^3+3ab^2C)+(b^3C+3a^2b)\sqrt{C}$$ $$\Rightarrow A+B\sqrt{C}=(a^3+3ab^2C)+(b^3C+3a^2b)\sqrt{C}$$ $$\Rightarrow \begin{cases} A=a^3+3ab^2C&(1) \\ B=b^3C+3a^2b&(2) \end{cases}$$ $$\text{Let}~\alpha=\frac{A}{B}\Rightarrow A=\alpha B.\text{ Multiply equation (2) by }\alpha\text{ and we get }$$ $$B\alpha=\alpha b^3C+3\alpha a^2b.\text{ But }B\alpha=A=a^3+3ab^2C$$ $$\Rightarrow a^3+3ab^2C=\alpha b^3C+3\alpha a^2b$$ $$\text{Dividing by }b^3\text{ we get }\Big(\frac{a}{b}\Big)^3+3C\Big(\frac{a}{b}\Big)=3\alpha\Big(\frac{a}{b}\Big)^3+\alpha C.$$ $$\text{ Make the notation }w=\frac{a}{b}\text{ so }w^3+3Cw=3\alpha w^3+\alpha C.$$ $$\text{Moving terms to the left hand side we have }\underbrace{1}_aw^3\underbrace{-3\alpha}_b w^2\underbrace{+3C}_cw\underbrace{-\alpha C}_d=0.$$ $$\text{Solve the cubic equation using the cubic formula. I have emphasized the coefficients above.}$$ $$\Delta_0=b^2-3ac=9(\alpha^2-C)$$ $$\Delta_1=2b^3-9abc+27a^2d=-54\alpha(\alpha^2-C)$$ $$\text{Now calculate the cubic constant }$$ $$\mathcal{C}=\sqrt[3]{\frac{\Delta_1\pm\sqrt{\Delta_1^2-4\Delta_0^3}}{2}}$$ $$\text{ Note this }\mathcal{C}\text{ is not the }C\text{ from the nested radical! }$$ $$\sqrt{\Delta_1^2-4\Delta_0^3}=\sqrt{\big(-54\alpha(\alpha^2-C)\big)^2-4\big(9(\alpha^2-C)\big)^3}=\sqrt{54^2(\alpha^2-C)^2\big(\alpha^2-(\alpha^2-C)\big)}$$ $$\sqrt{\Delta_1^2-4\Delta_0^3}=54|\alpha^2-C|\sqrt{C}$$ $$\mathcal{C}=\sqrt[3]{\frac{-54\alpha(\alpha^2-C)\pm54|\alpha^2-C|\sqrt{C}}{2}}$$ $$\text{But the }\pm\text{ and the }|\alpha^2-C|\text{ go meh and we are left with }$$ $$\mathcal{C}=\sqrt[3]{\frac{-54\alpha(\alpha^2-C)\pm54(\alpha^2-C)\sqrt{C}}{2}}$$ $$\text{Let's consider the }-\text{ solution because we will have}-54\text{ common factor and we will be left only with }+\text{ inside the radical. }\mathcal{C}\text{ simplifies to be}$$ $$\mathcal{C}=\sqrt[3]{\frac{-54(\alpha^2-C)(\alpha-\sqrt{C})}{2}}=-3\sqrt[3]{(\alpha^2-C)(\alpha+\sqrt{C})}$$ $$\text{One of the roots will have the formula }w=-\frac{1}{3a}\Big(b+\mathcal{C}+\frac{\Delta_0}{\mathcal{C}}\Big).$$ $$\frac{\Delta_0}{\mathcal{C}}=\frac{9(\alpha^2-C)}{-3\sqrt[3]{(\alpha^2-C)(\alpha^2-\sqrt{C})}}=-3\frac{(\alpha+\sqrt{C})(\alpha-\sqrt{C})}{\sqrt[3]{\alpha+\sqrt{C}}\sqrt[3]{(\alpha-\sqrt{C})^2}}$$ $$\text{We have }\frac{z}{\sqrt[3]{z}}=\sqrt[3]{z^2}\text{ and }\frac{z}{\sqrt[3]{z^2}}=\sqrt[3]{z}\text{ by rationalizing.}$$ $$\text{For }z=\alpha+\sqrt{C}\text{ and }z=\alpha-\sqrt{C}\text{ respectively we obtain }$$ $$\frac{\Delta_0}{\mathcal{C}}=-3\sqrt[3]{(\alpha^2+C)(\alpha-\sqrt{C})}$$ $$w=-\frac{1}{3}\Big(-3\alpha-3\sqrt[3]{(\alpha^2-C)(\alpha-\sqrt{C})}-3\sqrt[3]{(\alpha^2-C)(\alpha+\sqrt{C})}\Big)$$ $$w=\alpha+\sqrt[3]{\alpha^2-C}\Big(\sqrt[3]{\alpha-\sqrt{C}}+\sqrt[3]{\alpha+\sqrt{C}}\Big)$$ $$\text{Make the following notations: } \begin{cases} R=\sqrt[3]{\alpha^2-C} \\ T=\underbrace{\sqrt[3]{\alpha-\sqrt{C}}}_p+\underbrace{\sqrt[3]{\alpha+\sqrt{C}}}_q \end{cases}$$ $$\text{We will work on }T.~T^3=(p+q)^3=p^3+q^3+3pq(p+q).$$ $$p^3+q^3=2\alpha$$ $$3pq(p+q)=3\sqrt[3]{\alpha^2-C}\cdot T=3RT$$ $$T^3=2\alpha+3RT\Rightarrow T^3-3RT-2\alpha=0$$ $$\mathbf{And~here~we~stop.}$$ This is a depressed equation which should be OK to solve by hand. We will suppose $$T$$ is calculated through this equation. Hence, we have calculated: $$\alpha$$, $$R$$ and $$T$$ and should be able to go for $$w=\alpha+RT$$. Coming back to our system of equations, we have $$\frac{a}{b}=w\Rightarrow a=wb$$, and by substituting in $$(2)$$ we get $$B=3w^2b^3+Cb^3\Rightarrow b^3(3w^2+C)=B\Rightarrow b=\sqrt[3]{\frac{B}{3w^2+C}}$$ $$B=b^3C+3a^2b\Rightarrow 3a^2b=B-b^3C\Rightarrow a=\pm\sqrt{\frac{B-b^3C}{3b}}$$ $$\mathbf{And~these~are~the~final~formulas.}$$ Let's recap the algorithm for $$\sqrt[3]{A+B\sqrt{C}}$$. We have $$\begin{cases} \alpha=\frac{A}{B} \\ R=\sqrt[3]{\alpha^2-C} \\ T^3-3RT-2\alpha=0 \\ w=\alpha+RT \\ b=\sqrt[3]{\frac{B}{3w^2+C}} \\ a=\pm\sqrt{\frac{B-b^3C}{3b}} \end{cases}$$ Let's take for example $$\sqrt[3]{2+\sqrt{5}}$$. $$\begin{cases} A=2 \\ B=1 \\ C=5 \\ \alpha=2 \\ R=-1 \\ T^3+3T-4=0\Rightarrow T=1\text{ (obvious)} \\ w=1 \\ b=\frac{1}{2} \\ a=\frac{1}{2} \end{cases}$$ Our final form is $$a+b\sqrt{C}=\frac{1}{2}+\frac{1}{2}\sqrt{5}$$. So $$\sqrt[3]{2+\sqrt{5}}=\frac{1}{2}+\frac{1}{2}\sqrt{5}$$.

Another example: $$\sqrt[3]{7+5\sqrt{2}}$$. $$\begin{cases} A=7 \\ B=5 \\ C=2 \\ \alpha=\frac{7}{5} \\ R=-\sqrt[3]{\frac{1}{25}}=-\frac{\sqrt[3]{5}}{5} \\ T^3+\frac{3\sqrt[3]{5}}{5}T-\frac{14}{5}=0\Leftrightarrow5T^3+3\sqrt[3]{5}T-14=0 \\ \end{cases}$$ $$T\text{ will be of the form }\frac{k}{\sqrt[3]{5}}\Rightarrow k^3+3k-14=0.\text{ By trial we get }k=2\text{ and }T=\frac{2}{\sqrt[3]{5}}$$ $$\begin{cases} w=1 \\ b=1 \\ a=1 \end{cases}$$ Conclusion: $$\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$$

Let's see the example from the question: $$\sqrt[3]{-27+6\sqrt{21}}$$ $$\begin{cases} A=-27 \\ B=6 \\ C=21 \\ \alpha=-\frac{9}{2} \\ R=-\frac{\sqrt[3]{3}}{\sqrt[3]{2}}=-\frac{\sqrt[3]{6}}{2} \\ T^3+\frac{3\sqrt[3]{6}}{2}T+9=0\Leftrightarrow2T^3+3\sqrt[3]{6}T+18=0 \\ \end{cases}$$ $$T\text{ will be of the form }\frac{k}{\sqrt[3]{6}}\Rightarrow\frac{k^3}{3}+3k+18=0\Leftrightarrow k^3+9k+54=0.$$ $$k\text{ must be negative.}$$ $$\text{For }-1\text{ and }-2\text{ we get }44\text{ and }28\text{ respectively.}$$ $$\text{It feels like we're approaching the answer. Let's try }-3\text{ and we get identity.}$$ $$k=-3\text{ with }T=-\frac{3}{\sqrt[3]{6}}$$ $$\begin{cases} w=-3 \\ b=\frac{1}{2} \\ a=-\frac{3}{2}\text{ (notice that we used the negative solution)} \end{cases}$$ Thus, $$\sqrt[3]{-27+6\sqrt{21}}=-\frac{3}{2}+\frac{\sqrt{21}}{2}$$.

FINAL NOTES

• I am a high school student. I have no real experience with math. I have no idea how correct these things are.
• I am aware there is a whole documentation on nested radicals (Galois theory), but it was fun deducing these formulas on my own.
• I couldn't find one, but there might be another simpler formula for this on the internet.
• If one were to seriously interpret this, I hope it's inspiration for future proofs and formulas.
• I would like to see criticism. It's my first time doing this kind of formula deducing.

Signed by Neox

• I am sorry for the bad formatting. In the preview it looked much more clear.
– Neox
Jun 22, 2021 at 20:58
• Please review carefully the current policy of Enforcement of Quality Standards, which now applies to not only askers, but also the answerers that answer very low quality questions. Jun 22, 2021 at 21:14
• I am well aware of the poor context of my question, Aaron Hendrickson. Although, I specificed that the idea originates from another post of mine, where I was asking for such a formula. There, I received a comment with a post where the denesting is solved for a particular radical, which inspired me to generalize it. I am kind of new to the forum and I didn't take my time to strictly read all the rules. Also, when I considered to make the post, I wouldn't categorize the idea as very low quality, since I've seen many other posts on this topic and to be honest they didn't seem too complicated.
– Neox
Jun 22, 2021 at 21:54
• I see just now that you both were the questioner and answerer. The motivation of my comment is that we have seen a great influx of low quality questions (usually HW problems with no attempt), followed up with answers. This creates a perverse incentive for more bad questions. But in this case since you performed both tasks (great answer btw) and so I'm not sure my earlier comment applies. Jun 22, 2021 at 21:59
• @AaronHendrickson I see now that you have misunderstood the post. Though I appreciate that you took your time to redirect me to the posting rules FAQ. See the good part - I've learned something new!
– Neox
Jun 22, 2021 at 22:44

Thr denest formula is

$$\sqrt[3]{A+B \sqrt C}=\frac12\sqrt[3]{3Bs+2A}\left(1+\frac {\sqrt C}{s+A/B}\right)$$ with $$s$$ satisfying the depressed cubic equation $$s^3+3(C-\frac{A^2}{B^2})s +\frac{2A}B (C-\frac{A^2}{B^2}) =0$$

For $$\sqrt[3]{-27+6\sqrt{21}}$$, one has $$s^3+\frac94s-\frac{27}4=0$$ and $$s=\frac32$$, leading to the denesting $$\sqrt[3]{-27+6\sqrt{21}} = \frac12\sqrt[3]{18s-54}\left(1+ \frac{\sqrt {21}}{s+ 7/2}\right) =-\frac32+\frac12\sqrt{21}$$

OP's example is one of the cases covered by my answer here.

Therefore 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 $$\,a,b = \frac{1}{2}\left(t'\,\sqrt{p} \pm \sqrt{t'^{\,2} p-4r}\right)\,$$.

For $$\,a,b = \sqrt[3]{-27 \pm 6\sqrt{21}}\,$$:

• $$m=-27\,$$, $$\,p=1\,$$, $$\,n=\pm 6\,$$, $$\,q=21\,$$;

• $$m^2 \cdot p - n^2 \cdot q$$ $$= (-27)^2 - (\pm 6)^2 \cdot 21$$ $$= -27$$ $$=(-3)^3$$ $$\implies r=-3\,$$;

• $$0 = p\, t'^{\,3} - 3r\, t' - 2m = t'^{\,3} + 9 t' + 54\,$$ with the only real root $$\,t' = -3\,$$.

Then:

$$a,b = \frac{1}{2}\left(t'\,\sqrt{p} \pm \sqrt{t'^{\,2} p - 4r}\right) = \frac{1}{2}\left(-3 \cdot 1 \pm \sqrt{(-3)^2 \cdot 1 - 4 \cdot (-3)}\right) = \frac{1}{2}\left(-3 \pm \sqrt{21}\right)$$