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I was reading a pdf on Cardano's method of solving for the roots of a cubic polynomial, when I noticed an example. $$x^3+6x-20=0$$ On solving, I got $$x=\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}$$ I went through a calculator and it gave the answer $2$. My question is how can we prove $$\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}=2$$ I tried to simplify this radical, but arrive at the same cubic polynomial. No further advancement is taking place in my solving. Another similar interesting cubic polynomial is $$x^3-15x-4=0$$ It has the root $$x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}$$ and read that $$\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}=4$$ I hope someone will help me.

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As $\displaystyle10+\sqrt{108}=10+6\sqrt3,$

we can write $\displaystyle10+6\sqrt3=(a+b\sqrt3)^3$ where $a,b$ are real rationals

$\displaystyle\implies10+6\sqrt3=a^3+2a^2\cdot b\sqrt3+3a(b\sqrt3)^2+(b\sqrt3)^3$ $\displaystyle\implies10+6\sqrt3= a^3+9ab^2+3(a^2b+b^3)\sqrt3$

Comparing the rational & the irrational parts $\displaystyle 10=a^3+9ab^2\ \ \ \ (1)$ and $\displaystyle a^2b+b^3=2$

$\displaystyle\implies5(a^2b+b^3)=a^3+9ab^2\iff a^3-5a^2b+9ab^2-5b^3=0$

Clearly, $a=b$ is a solution

From $\displaystyle(1),10a^3=10\iff a^3=1\iff a=1$ as $a$ is real

So, from $(1)b^2=1$ and from $(2),b^3+b-2=0$ the common root being $1$

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  • $\begingroup$ @BalarkaSen, We are lucky that $a=b$ is so evident:) $\endgroup$ Dec 27, 2013 at 15:34
  • $\begingroup$ @BalarkaSen, do $a,b$ need to be integers? Else how shall we iterate ? $\endgroup$ Dec 27, 2013 at 15:42
  • $\begingroup$ @BalarkaSen, I assumed them to be rationals only, but how to prove that they must be integers? $\endgroup$ Dec 27, 2013 at 15:46
  • $\begingroup$ thid was what i needed.accepted $\endgroup$
    – Suraj M S
    Dec 27, 2013 at 17:14
  • $\begingroup$ @SurajM.S, nice to hear that . Related : math.stackexchange.com/questions/386488/… $\endgroup$ Dec 28, 2013 at 17:00
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Another way to find out the solutions is the following. For the fundamental theorem of algebra the solutions are three. You can easily see that $x^3+6x-20=(x-2)(x^2+2x+10)$ than there is one real solution $x_1=2$. The other two solutions are complex, in accordance with the Cardano's formulas

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The following way does show that your value equals $2$.

Letting $f(x)=x^3+6x-20,$ since we have $$f^\prime(x)=3x^2+6\gt0.$$ Hence, $f(x)=0$ has exactly one real solution.

Hence, we know that the following holds :

$$\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}=2.$$

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  • $\begingroup$ I know it. But you'll find a sentence as "my question is how can we prove...". My answer is one of the proofs. $\endgroup$
    – mathlove
    Dec 27, 2013 at 15:09
  • $\begingroup$ I totally agree with you. $\endgroup$
    – mathlove
    Dec 27, 2013 at 15:15
  • $\begingroup$ can you explain some more how this can be used to prove the value is $2$ $\endgroup$
    – Suraj M S
    Dec 27, 2013 at 15:26
  • $\begingroup$ Well, you found the value is a solution of the given equation. Then, you know that the value is a real number. By the way, we know that $y=f(x)$ is a strictly monotone increasing function. This means that the graph has only one intersection point of $x$-axis. So, we know that $f(x)=0$ has only one real solution. Since $f(2)=0$, we know $2$ is a solution. Hence, your value has to be equal to $2$ because there is only one real solution! $\endgroup$
    – mathlove
    Dec 27, 2013 at 15:32
  • $\begingroup$ @BalarkaSen: Oh, nice point. Thank you. $\endgroup$
    – mathlove
    Dec 27, 2013 at 15:36

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