# Why am i getting two real roots of the cubic $x^3+x+5=0$

I was going through the solution of depressed cubic equation $$x^3+x+5=0$$

By Cardano's Method we assume

$$x=u+v$$ Then

we have $$(u+v)^3-3uv(u+v)-(u^3+v^3)=0$$ Comparing with given cubic we get

$$u^3+v^3=-5 \tag{1}$$

$$uv=\frac{-1}{3}\tag{2}$$ Solving $$(1)$$ and $$(2)$$ we get

$$u^3-\frac{1}{27u^3}=-5$$ Letting $$u^3=p$$ we get

$$27p^2+135p-1=0$$ which gives two real distinct roots of $$p$$ which $$\implies$$ two real distinct roots of $$u$$ and hence two real distinct roots of $$v$$

Hence two distinct real roots of $$x$$

what went wrong here?

• Did you note that what you wrote occurs when you apply Cardano's method to any reduced cubic? – José Carlos Santos Nov 24 '18 at 19:55

You will get two solutions because both $$u^3$$ and $$v^3$$ are solutions of the quadratic, and the second solution simply exchanges the two. The two solutions for the quadratic therefore give the same solutions for the cubic (depending which cube root you take).