Find the roots of $z^3 +3z^2 +3z +3=0$

Question

Hello I have this problem:

Find all $z \in C$

$z^3 +3z^2 +3z +3=0.$

With Mathematica I get 3 roots

$z_1 = -1 -\sqrt[3]{2}$

$z_2 = -1 + (1 + i \sqrt{3})/2^{(2/3)}$

$z_3 = -1 + (1 - i \sqrt{3})/2^{(2/3)}$

But I don't know how can find this roots. Also I don't know to reduce this cubic equation to a quadratic equation

How do you solve this?

Any help will be appreciated

Thanks you

Answer 1

Note that$$z^3+3z^2+3z+3=0\iff(z+1)^3=-2.$$Can you take it from here?

Answer 2

Now that you know the roots, it’s easy to express the cubic as the product of linear and quadratic terms. The cubic is just $(z-z_1)(z-z_2)(z-z_3)$, and the second two factors combine to give you a quadratic.

If you don’t already have the roots (which would be the usual situation), you have to use a bit of creativity, as in the answer from José, or you have to use one of the cubic roots formulae.

Answer 3

To solve a cubic of the form $z^3+bz^2+cz+d=0$, you can make the substitution $y=z+b/3$ to get a cubic in $y$ where the coefficient of $y^2$ is zero; this is known as a "depressed cubic". To solve a depressed cubic equation, you would usually make another substitution (e.g Vieta's substitution), which leaves you with a quadratic equation. However, in this case it is even simpler: by setting $y=z+1$, we see that $$ (y-1)^3+3(y-1)^2+3(y-1)+3=y^3+2=0 \, . $$ Hence, $y$ equals one of the three cube roots of $-2$, and so $z$ equals $-1$ plus one of the cube roots of $-2$.