### Instruction: Solve
\[(x^3 + 3x^2 \sqrt{2} + 6x + 2 \sqrt{2}) + (x + \sqrt{2}) = 0.\]Enter all the solutions, separated by commas.### Response: This looks like a cubic equation, but with some square roots of 2 involved.
I wonder if I can factor it somehow, or use some kind of substitution to make it easier.
I notice that the second term in each parenthesis has a factor of $\sqrt{2}$, and that the first term in each parenthesis is a multiple of $x$.
Maybe I can use that to simplify things.
Let me try letting $y = x + \sqrt{2}$.
Then $x = y - \sqrt{2}$, and I can plug that into the equation.
I get \[(y - \sqrt{2})^3 + 3(y - \sqrt{2})^2 \sqrt{2} + 6(y - \sqrt{2}) + 2 \sqrt{2} + y = 0.\]
This still looks messy, but maybe I can expand it and see what happens.
Expanding the cubes and squares, I get \[y^3 - 3y^2 \sqrt{2} + 6y - 6 \sqrt{2} + 3y^2 \sqrt{2} - 12y + 12 + 6y - 6 \sqrt{2} + 2 \sqrt{2} + y = 0.\]
Now I can simplify by combining like terms.
The terms with $\sqrt{2}$ cancel out, and I'm left with \[y^3 + y = 0.\]
Wow, that's much nicer.
I can factor out a $y$ and get \[y(y^2 + 1) = 0.\]
Now I can use the zero product property and set each factor equal to zero.
For $y = 0$, I get $x = y - \sqrt{2} = -\sqrt{2}$ as one solution.
For $y^2 + 1 = 0$, I get $y^2 = -1$, which means $y = \pm i$, where $i$ is the imaginary unit.
Then $x = y - \sqrt{2} = \pm i - \sqrt{2}$ as the other two solutions.
So I have three solutions, and I can write them in the form asked.