### Instruction: Solve
\[(x^3 + 3x^2 \sqrt{2} + 6x + 2 \sqrt{2}) + (x + \sqrt{2}) = 0.\]Enter all the solutions, separated by commas.### Response: To solve this equation, I will try to factor the left-hand side using the sum of cubes formula.
I notice that the first term is a perfect cube, since $x^3 = (x)^3$.
I also notice that the last term is a perfect cube, since $2\sqrt{2} = (\sqrt{2})^3$.
So I can write the left-hand side as $(x + \sqrt{2})^3$.
Now I can use the sum of cubes formula, which says that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, where $a = x + \sqrt{2}$ and $b = x + \sqrt{2}$.
So I get $(x + \sqrt{2})^3 + (x + \sqrt{2}) = (x + \sqrt{2})(x + \sqrt{2})^2 + (x + \sqrt{2}) = 0$.
Now I can factor out $(x + \sqrt{2})$ from both terms and get $(x + \sqrt{2})((x + \sqrt{2})^2 + 1) = 0$.
To find the solutions, I need to set each factor equal to zero and solve for x.
The first factor gives me $x + \sqrt{2} = 0$, which means $x = -\sqrt{2}$.
The second factor gives me $(x + \sqrt{2})^2 + 1 = 0$, which means $(x + \sqrt{2})^2 = -1$.
To solve for x, I need to take the square root of both sides, but I have to remember that there are two possible values for the square root of -1: $i$ and $-i$, where $i$ is the imaginary unit.
So I get $x + \sqrt{2} = i$ or $x + \sqrt{2} = -i$.
Solving for x, I get $x = i - \sqrt{2}$ or $x = -i - \sqrt{2}$.