Cho \( a + b + c = 0 \), Rút gọn \[ abc = a^3 + b^3 + c(a^2 + b^2) - abc \]

Given the condition \( a + b + c = 0 \), we want to simplify the expression:

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

We can rearrange it as follows:

\[
abc + abc = a^3 + b^3 + c(a^2 + b^2).
\]

This simplifies to:

\[
2abc = a^3 + b^3 + c(a^2 + b^2).
\]

Next, we can use the identity for the sum of cubes:

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

Since \( a + b + c = 0 \), it follows that:

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

This means we can express \( c \) in terms of \( a \) and \( b \) as \( c = -a - b \). Substituting \( c \) in the expression for \( a^3 + b^3 + c(a^2 + b^2) \):

\[
a^3 + b^3 + (-a-b)(a^2 + b^2).
\]

Now, we can expand:

\[
a^3 + b^3 - (a + b)(a^2 + b^2) = a^3 + b^3 - a^3 - ab^2 - b^3 - a^2b = - ab(a + b).
\]

Thus, we have:

\[
2abc = - ab(a + b).
\]

Since \( a + b = -c \), we obtain:

\[
2abc = ab(-c).
\]

Dividing both sides by \( ab \) (assuming \( ab \neq 0 \)), we find:

\[
2c = -c \implies 3c = 0 \implies c = 0.
\]

From this, we have \( a + b = 0 \), implying \( b = -a \).

Substituting back, we can now simplify the expression fully under the premise \( c = 0 \):

\[
abc = a^3 + (-a)^3 + 0 = a^3 - a^3 = 0.
\]

Thus, combining all these results, we have:

\[
abc = 0.
\]

In summary, the final simplification results in:

\[
abc = 0.
\]