How do you factor $6{x^3} + 6 = 0$?

Hint: In this question, we have been asked to factorize a cubic equation. But the given equation is not similar to the standard cubic equation $\left( {a{x^3} + b{x^2} + cx + d = 0} \right)$. So, how to solve this equation? First, take out the constant as it is common in both the terms. Then, you will observe a cubic equation. Open it using the cubic formula. After applying the formula, you will get all the factors.

Formula used:
${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$

Complete step-by-step answer:
We are given a cubic equation and we have been asked to factorize it. Let us see how we can do it.
$ \Rightarrow 6{x^3} + 6 = 0$ …. (given)
Taking $6$ common out of the terms,
$ \Rightarrow 6\left( {{x^3} + 1} \right) = 0$
We can also write it as –
$ \Rightarrow 6\left( {{x^3} + {1^3}} \right) = 0$
Now, we will use the cubic formula to expand and find our factors.

$ \Rightarrow 6\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) = 0$
Hence, these are the factors of the given cubic equation.
One thing to be noted is that the number of factors/solutions is always equal to the degree of the question.

Note:
Children often find the factors in a wrong way. Let us see how it should not be done.
We are given $6{x^3} + 6 = 0$.
Shifting the constant term to the other side
$ \Rightarrow 6{x^3} = - 6$
Simplifying the term,
$ \Rightarrow {x^3} = \dfrac{{ - 6}}{6} = - 1$
Now, in order to solve this, students again shift the constant term to the other side.
$ \Rightarrow {x^3} + 1 = 0$
After this step, students apply the identity of ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$.
What is wrong in this? As mentioned above, the number of factors/solutions is always equal to the degree of the question. If we use this identity now, I have already lost one factor, i.e., $6$.
So, you should never cancel out the terms in a question where you have been asked to factorise.