Given the equation [tex]\(x + y + z = 0\)[/tex], we want to prove that [tex]\(x^3 + y^3 + z^3 = 3xyz\)[/tex].
Let's go through the steps methodically to derive the required result.
1. Use an identity for the sum of cubes: We know that in algebra, the sum of cubes can be represented by the following identity:
[tex]\[
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
\][/tex]
2. Substitute [tex]\(x + y + z = 0\)[/tex] into the identity: Given that [tex]\(x + y + z = 0\)[/tex], substitute this into the identity:
[tex]\[
x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)
\][/tex]
3. Simplify the equation: The right-hand side of the equation becomes zero:
[tex]\[
x^3 + y^3 + z^3 - 3xyz = 0
\][/tex]
Therefore:
[tex]\[
x^3 + y^3 + z^3 = 3xyz
\][/tex]
Thus, we have derived that:
[tex]\[
x^3 + y^3 + z^3 = 3xyz
\][/tex]
This completes the proof, demonstrating that under the condition [tex]\(x + y + z = 0\)[/tex], the equation [tex]\(x^3 + y^3 + z^3\)[/tex] indeed equals [tex]\(3xyz\)[/tex].
