algebra precalculus - Show that $frac {37 - sqrt{1357}}{3} = frac {4}{37 + sqrt{1357}}$.

Solving the cubic equation:

$$3x^3 - 74x^2 + 4x = 0$$

I found the following roots:

$$x = 0, \frac {37 - \sqrt{1357}}{3}, \frac {37 + \sqrt{1357}}{3}$$

On the Wolfram Alpha website one of the roots is shown differently.

Instead of the root:

$$x = \frac {37 - \sqrt{1357}}{3}$$

The site shows:

$$x = \frac {4}{37 + \sqrt{1357}}$$

I rapidly established that:

$$\frac {37 - \sqrt{1357}}{3} = \frac {4}{37 + \sqrt{1357}} \approx 0.054$$

But I can't see what mathematical steps would turn $\frac {37 - \sqrt{1357}}{3}$ into $\frac {4}{37 + \sqrt{1357}}$ or indeed visa-versa. Can someone explain what steps would make this transformation please?

Also...

It seems to me that if an equation has 2 roots which differ only in whether they have a plus or minus sign as a result of taking the square root of both sides of a quadratic equation, then it's best to express the 2 roots in the same way with only the plus-or-minus sign differing (it's clearer that way right?). What are the reasons for expressing one of them differently as Wolfram Alpha has done in this case?

Thanks.

Answer

$$\frac {37 - \sqrt{1357}}{3}= \frac{(37-\sqrt{1357})(37+\sqrt{1357})}{3(37+\sqrt{1357})}=\frac{37^2-1357}{3(37+\sqrt{1357})}$$ $$=\frac{12}{3(37+\sqrt{1357})}=\frac{4}{(37+\sqrt{1357})}$$