Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$

Question

Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$

The way I attempted to the question is the following:

I state that $a=\sqrt[3]{x+1}$ hence we have that $3a^3-3+a+a^2=-1$. Hence $3a^3-2+a+a^2=0$, so $a^2(3a+1)+a-3=0$. And this is where I got stuck.

Could you please explain to me how to finish off the question with this train of thought and out of 10, what would be logical to assume that I would get for these workings out?

Answer 1

$3a^3+a^2+a-2=0$

$(3a-2)(a^2+a+1)=0$

We can find this by using Rational Root theorem, or another way is by noting that $a^2+a+1$ is also a root because $a^2+a+1=0 \implies a^3=1$ (roots of unity) so $$3a^3+a^2+a-2=3+a^2+a-2=a^2+a+1=0$$

Answer 2

$$ 3 a^3-2+a+a^2=(3 a - 2) (a^2 + a + 1) $$