A radical idea

**NowIsForever** · May 15th 2010, 06:52 PM

My source said that someone "was fooling around with Cardan's formula for solving x^3 + 6x = 20 and came up with (10 + sqrt(108))^(1/3) + (10 - sqrt(108))^(1/3) as a solution. Of course that has to equal 2, but the question is 'how do we know'? If we simply cube each of the radicals, we end up eventually with the original equation."

And my question is: Can you prove this?

(I have a simple proof which I'll post if no one does so first.)

**running-gag** · May 16th 2010, 12:26 AM

Hi

As you said the equation $x^3+6x-20=0$ has one obvious solution, which is 2
It is the only real solution since $x^3+6x-20=(x-2)(x^2+2x+10)$ and the discriminant of the quadratic is negative

Using Cardan formula gives a solution as $\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}$

Of course since 2 is the only real solution this expression is equal to 2

This is a proof

**HallsofIvy** · May 16th 2010, 04:43 AM

Way to go, running-gag! Although I don't think that was quite what NowisForever had in mind.

NowisForever, here's the hard way:
$(a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$ so, in particular, $x^3= \left(\sqrt[3]{10+ \sqrt{108}}+ \sqrt[3]{10- \sqrt{108}}\right)^3$ $= 10+ \sqrt{108}+ 3\sqrt[3]{10+\sqrt{108})^2(10- \sqrt{108}}$ $+ 3\sqrt[3]{(10+ \sqrt{108})(10- \sqrt{108})^2+ 10- \sqrt{108}}$

Now, $(10+ \sqrt{108})^2(10- \sqrt{108})= (10+ \sqrt{108})(10+ \sqrt{108})(10- \sqrt{108})$ $= (10+ \sqrt{108})(100- 108)= -8(10+ \sqrt{108}$ and, equivalently, $(10+ \sqrt{108})(10- \sqrt{108})^2= -8(10-\sqrt{108}$ .

So that cube becomes $x^3= 20- 6\sqrt[3]{10+ \sqrt{108}}- 6\sqrt[3]{10- \sqrt{108}}$ .

But $6x= 6\sqrt[3]{10+ \sqrt{108}}+ 6\sqrt[3]{10- \sqrt{108}}$ so that $x^3+ 6x= 20$ and the equation follows.

**NowIsForever** · May 16th 2010, 07:38 AM

FWIW, this is what I had in mind:

10 + 6*3^(1/2) = 1 + 3*3^(1/2) + 3*3 + 3*3^(1/2)
(10 + 6*3^(1/2))^(1/3) = 1 + 3^(1/2)

10 - 6*3^(1/2) = 1 - 3*3^(1/2) + 3*3 - 3*3^(1/2)
(10 - 6*3^(1/2))^(1/3) = 1 - 3^(1/2)

(10 + sqrt(108))^(1/3) + (10 - sqrt(108))^(1/3) =
(10 + 6*3^(1/2))^(1/3) + (10 - 6*3^(1/2))^(1/3) = 1 + 3^(1/2) + 1 - 3^(1/2) = 2

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