#### NowIsForever

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

Hi

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

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

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

This is a proof

HallsofIvy

#### HallsofIvy

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

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

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

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

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

running-gag

#### NowIsForever

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