### Instruction: Suppose that there exist nonzero complex numbers $a,$ $b,$ $c,$ and $d$ such that $k$ is a root of both the equations $ax^3 + bx^2 + cx + d = 0$ and $bx^3 + cx^2 + dx + a = 0.$  Enter all possible values of $k,$ separated by commas.### Response: The problem asks me to find all possible values of $k$ that are roots of both equations, which means they make both equations true when I plug them in for $x.$
I could try to solve each equation separately, but that might be messy and time-consuming, especially since they involve complex numbers.
Maybe there is a clever way to use the fact that the coefficients of the two equations are related by a cyclic permutation.
I notice that if I multiply the first equation by $x$ and subtract the second equation from it, I get $x^4 - 1 = 0.$
This is a much simpler equation to solve, and any root of this equation must also be a root of both original equations, since it is a linear combination of them.
To solve $x^4 - 1 = 0,$ I can factor it as $(x^2 + 1)(x^2 - 1) = 0.$
Then, using the difference of squares formula, I can further factor it as $(x + i)(x - i)(x + 1)(x - 1) = 0.$
Therefore, the only possible values of $k$ that satisfy the equation are $i, -i, 1,$ and $-1.$
These are the fourth roots of unity, and they have the property that any cyclic permutation of the coefficients of $x^3 + x^2 + x + 1 = 0$ leaves the equation unchanged.
To check that these values are indeed roots of both original equations, I can plug them in and simplify.
For example, if $k = i,$ then the first equation becomes $a(i^3) + b(i^2) + c(i) + d = 0,$ which simplifies to $-ai - b + ci + d = 0.$
The second equation becomes $b(i^3) + c(i^2) + d(i) + a = 0,$ which simplifies to the same expression.
Therefore, $i$ is a root of both equations.