Why does Wolfram Alpha say the roots of a cubic involve square roots of negative numbers, when all three roots are real?

This is an example of the casus irreducibilis: the formulas for solving cubics need complex numbers when the cubic has three real roots.

When the discriminant of the equation is negative, there are $3$ real roots. It's precisely in this case that Cardano's formulæ do not work, because the radicands are negative.

It is even historically the reason why square roots of negative numbers were introduced: the formulae become valid in all cases.

For your interest, the real representation of the roots is given by

$$x_k=2\cos\left(\frac\pi9+\frac{2\pi k}3\right)\qquad k=0,1,2$$

I always find these trigonometric representations nicer.

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This comes directly from the following identity:

$$\cos(3a)=4\cos^3(a)-3\cos(a)$$

$$\cos(3\arccos(a))=4a^3-3a$$

And lastly the fact that cosine is periodic.