Let $a,$ $b,$ $c$ be non-zero real numbers such that $a + b + c = 0.$  Find all possible values of
\[\frac{a^3 + b^3 + c^3}{abc}.\]Enter all the possible values, separated by commas.
From the equation $a + b + c = 0,$ $c = -a - b.$  Hence,
\begin{align*}
\frac{a^3 + b^3 + c^3}{abc} &= -\frac{a^3 + b^3 - (a + b)^3}{ab(a + b)} \\
&= \frac{3a^2 b + 3ab^2}{ab(a + b)} \\
&= \frac{3ab(a + b)}{ab(a + b)} \\
&= \boxed{3}.
\end{align*}By the Multivariable Factor Theorem, this implies that $a + b + c$ is a factor of $a^3 + b^3 + c^3 - 3abc.$  We can then factor, to get the factorization.
\[a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc).\]
3
Level 4
Intermediate Algebra
Problem: Let $a,$ $b,$ $c$ be non-zero real numbers such that $a + b + c = 0.$  Find all possible values of
\[\frac{a^3 + b^3 + c^3}{abc}.\]Enter all the possible values, separated by commas.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \(\frac{a^3 + b^3 + c^3}{abc}\), we can use the identity for the sum of cubes. The identity states that for any three numbers \(x, y, z\):

\[
x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz).
\]

In our case, let \(x = a\), \(y = b\), and \(z = c\). Given that \(a + b + c = 0\), we can substitute this into the identity:

\[
a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc).
\]

Since \(a + b + c = 0\), the equation simplifies to:

\[
a^3 + b^3 + c^3 - 3abc = 0.
\]

This implies:

\[
a^3 + b^3 + c^3 = 3abc.
\]

Now, substituting this result into our original expression, we have:

\[
\frac{a^3 + b^3 + c^3}{abc} = \frac{3abc}{abc} = 3.
\]

Thus, the only possible value for \(\frac{a^3 + b^3 + c^3}{abc}\) is \(3\).<|END_OF_LATENT|> The final answer is $3$.