$$\sqrt{A\pm\sqrt{B}}=\sqrt{\frac{A+C}{2}}\pm\sqrt{\frac{A-C}{2}}$$ , where $C^2=A^2-B$. But, I couldn't find a formula for $\sqrt[3]{A\pm\sqrt{B}}$.

First of all, why would you even need this? I've recently learned the cubic formula from Wikipedia. Let's consider the general cubic equation $ax^3+bx^2+cx+d=0$. We have $$\Delta_0=b^2-3ac$$ $$\Delta_1=2b^3-9abc+27a^2d$$ Then we calculate $$C=\sqrt[3]{\frac{\Delta_1\pm\sqrt{\Delta_1^2-4\Delta_0^3}}{2}}$$ Whether we chose $+$ or $-$ is trivial because we are anyways going to get the same roots, regardless the order.

Then we have the final formula: $$x_k=-\frac{1}{3a}\Big(b+C\omega^k+\frac{\Delta_0}{C\omega^k}\Big)$$ where $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ (primitive cube root of unity) and $x \in \{0,1,2\}$.

I prefer to calculate the first 2 roots using the formula and the last root from Vieta's formula $\sum x_k=-\frac{b}{a}$.

Let' take, for example, the following equation: $$x^3+x-2=0$$ with $$a=1,~b=0,~c=1,~d=-2$$ We have $$\Delta_0=-3$$ $$\Delta_1=-54$$ And C simplifies to be $$C=\sqrt[3]{-27+6\sqrt{21}}$$ I knew for sure the answer to this would be something of the form $A+\sqrt{B}$ , with $A,~B\in\mathbb{R}$. So $-27+6\sqrt{21}$ must be of the form $(x+y)^3$ with $x\in\mathbb{Q}$ and $y\in\mathbb{I}$ (it's something with $\sqrt{21}$). Expanding the expression we get $x^3+3x^2y+3xy^2+y^3=-27+6\sqrt{21}$. In this situation we can affirm that $x^3+3xy^2\in\mathbb{Q}$ and $y^3+3x^2y\in\mathbb{I}$. Hence, we have the following system of equations $$x^3+3xy^2=-27$$ $$y^3+3x^2y=6\sqrt{21}$$ I tried my guesses but nope, I couldn't find the solutions. I gave up and used WolframAlpha which immediately calculated it as $\frac{-3+\sqrt{21}}{2}$. So $x$ would be $-\frac{3}{2}$ and $y$ would be $\frac{\sqrt{3}}{2}$. I would have never thought of it. And solving the system by substitution, we have to substitute $y=\pm\sqrt{\frac{-x^3-27}{3x}}$ and that would lead to a monster equation, probably a hexic equation.

My question - is there a formula like the one for simple square roots? I couldn't find anything on the internet...

  • 1
    $\begingroup$ See Denesting Phi, Denesting Cube Roots for denesting $\sqrt[n]{A+B\sqrt[m]C}$. $\endgroup$
    – Vepir
    Commented Jun 22, 2021 at 13:07
  • $\begingroup$ Thanks, I understand it! Though, you have to solve another cubic equation to get to the answer :( $\endgroup$
    – Neox
    Commented Jun 22, 2021 at 13:25
  • $\begingroup$ See math.stackexchange.com/q/4148105. In fact you have to solve another cubic equation and this involves finitely many trials - but it always works. $\endgroup$
    – Paul Frost
    Commented Jun 22, 2021 at 14:21


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