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After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$ After plugging into the formula $$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}$$ where $m=6$ and $n=20$, we get $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}.$$ However, we notice that, without using Cardano's formula, that $x=2$ is the solution for the equation $x^3+6x=20.$ My question is: how does the equation $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}$$ get simplified to $x=2$?

P.S. I understand that it was Niccolo Fontana who first figured out how to solve depressed cubic, to give one the proper credit.

3 Answers3

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With the benefit of hindsight we notice that $10+\sqrt{108}=10+6\sqrt{3}$ and $$10+6\sqrt{3}=(1+\sqrt{3})^3.$$ Similarly, $$-10+6\sqrt{3}=(-1+\sqrt{3})^3.$$ Take the (real) cube roots and subtract.

André Nicolas
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  • Thanks Nicolas. How did you come up with the quantity $ \pm 1 + \sqrt{3}$? Surely you did not think of that quantity of blue, did you? –  May 10 '16 at 23:01
  • More or less yes. If there were going to be "nice" cube roots they would have to look like $\pm 1+r\sqrt{3}$, where $r$ is rational, because of the root $2$ that you observed. Now it takes little to see that $r=1$ works. – André Nicolas May 10 '16 at 23:04
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\begin{align*} &\left(10+\sqrt{108}\right)^{1/3}-\left(-10+\sqrt{108}\right)^{1/3}\\ &=\left(10+6\sqrt{3}\right)^{1/3}-\left(-10+6\sqrt{3}\right)^{1/3}\\ &=\left((1+\sqrt{3})^3\right)^{1/3}-\left((\sqrt{3}-1)^3\right)^{1/3}\\ &=(1+\sqrt{3})-(\sqrt{3}-1)\\ &=2 \end{align*}

The formula: $$(A+B\sqrt{3})^3=A^3+3\sqrt{3}A^2B+9B^2A+3\sqrt{3}B^3$$ is useful.

Siong Thye Goh
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Let $a^3 = 10+\sqrt{108}=10+6\sqrt{3}, b^3 = -10 + \sqrt{108}=-10+6\sqrt{3}\implies a^3-b^3 = 20 = (a-b)(a^2+ab+b^2)=(a-b)((a-b)^2 + 3ab)$. Observe that $(ab)^3 = 8 \implies ab = 2\implies (a-b)((a-b)^2 + 6) = 20\implies x^3+6x-20 = 0, x = a-b$. Using synthetic division we have: $x^3+6x-20 = 0\implies (x-2)((x+1)^2 + 9) = 0 \implies x = 2$ as claimed.

DeepSea
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