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Is there a way to find the roots of equations such as $x^3-9\sqrt[3]{2}+9=0$?

I've just been using Wolfram Alpha to factor it into $(x-\sqrt[3]{4}+\sqrt[3]{2}-1)(x^2+(1-\sqrt[3]{2}+\sqrt[3]{4})x+3\sqrt[3]{4}-3)$. But for harder equations such as $$x^3-63\sqrt[3]{20}+9=0$$, Wolfram Alpha just factors it into $$(x-\text{root of }x^9+27x^6+243x^3-5000211)(x-\text{same thing})(x-\text{same thing})$$ Which is kind of problematic, because I would like the exact values of the factors.

So is there a way to factor such polynomials into multiple factors? Or some way to find its roots! Anything helps.

Note: I would not like nested radicals such as $x=\sqrt[3]{3-\sqrt[4]{4}}$ because I then have to go through the process of finding there simplified surds.

Frank
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3 Answers3

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Say we have the polynomial $$x^3-(a\sqrt[3]b+c)=0$$ This is going to have one real root: $$x=\sqrt[3]{a\sqrt[3]b+c}$$ But you said you do not want nested radicals. Then we have the question of whether we can denest the radical.

For your first example with $x=\sqrt[3]{9\sqrt[3]2-9}$, you got lucky; the decomposition you gave was shown by Ramanujan. For your second example with $x=\sqrt[3]{63\sqrt[3]{20}-9}$, however, there is no solution with rational numbers and non-nested radicals only. There is an algorithm to determine whether or not the denesting can be done, and this MSE post has more on it (in general, it can't be done).

If a denesting does exist, the entire polynomial can be factored into expressions not involving nested radicals. If it doesn't exist, the nested radical provides the only exact solution; you could numerically compute $\sqrt[3]{a\sqrt[3]b+c}$ but it isn't going to be exact.

Parcly Taxel
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We can find $\sqrt[3]{9(\sqrt[3]2-1)}$ by the following way without WA.

Indeed, let $\sqrt[3]2=x$.

Hence, $$x^3=2$$ or $$9(x^3-1)=9$$ or $$9(x-1)=\frac{9}{1+x+x^2}$$ or $$9(x-1)=\frac{27}{x^3+3x^2+3x+1}$$ or $$\sqrt[3]{9(\sqrt[3]2-1)}=\frac{3}{\sqrt[3]2+1}$$ or $$\sqrt[3]{9(\sqrt[3]2-1)}=\sqrt[3]4-\sqrt[3]2+1$$ and we are done!

Michael Rozenberg
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  • Why did you start off with $x^3=2$? And how did you get from $$9(x-1)=\frac 9{1+x+x^2}$$to$$9(x-1)=\frac {27}{x^3+3x^2+3x+1}$$ – Frank Nov 20 '16 at 21:13
  • @Frank Because it's comfortable. $\frac{9}{1+x+x^2}=\frac{27}{3+3x+3x^2}=\frac{27}{1+x^3+3x+3x^2}$ – Michael Rozenberg Nov 20 '16 at 21:34
  • Wait, we have$$\frac {27}{3+3x+3x^2}=\frac {27}{1+x^3+3x+3x^2}$$So how would we work on the other one? I let $x=7\sqrt[3]{20}$ and along the way, I get $9(x^3-1)=6859\implies 9(x-1)=\frac {6859}{x^2+x+1}$. Now, I'm not sure what to do – Frank Nov 20 '16 at 21:44
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We can prove that $\sqrt[3]{63\sqrt[3]{20}-9}=\sqrt[3]{100}+\sqrt[3]{16}-\sqrt[3]5$

See here my generalization: Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$

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Michael Rozenberg
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