Given: x + y + z = 0

To Prove: x³ + y³ + z³ = 3xyz

Proof:
1. Introduction:
We are given that the sum of x, y, and z is equal to 0. We need to prove that the sum of their cubes, x³ + y³ + z³, is equal to 3 times their product, 3xyz.

2. Expanding the Cubes:
Let's expand the cubes of x, y, and z using the formula (a + b)³ = a³ + 3a²b + 3ab² + b³:

x³ + y³ + z³ = (x + y + z)(x² + y² + z² - xy - xz - yz)

3. Substituting the Given Condition:
We are given that x + y + z = 0. Substituting this in the above equation:

x³ + y³ + z³ = (0)(x² + y² + z² - xy - xz - yz)
x³ + y³ + z³ = 0

4. Proving 3xyz:
Now, let's prove that 3xyz is equal to 0 as well:

3xyz = 3(xy)(z) = 3(-z)(z) = -3z²

5. Comparing x³ + y³ + z³ and 3xyz:
We have x³ + y³ + z³ = 0 and 3xyz = -3z². Since we know that x + y + z = 0, we can substitute z as -x - y in 3xyz:

3xyz = -3(-x - y)² = -3(x² + y² + 2xy) = -3x² - 3y² - 6xy

6. Simplifying the Comparison:
Comparing x³ + y³ + z³ = 0 and 3xyz = -3x² - 3y² - 6xy, we can see that they are equal:

0 = -3x² - 3y² - 6xy

7. Conclusion:
Therefore, we have proved that if x + y + z = 0, then x³ + y³ + z³ = 3xyz.

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In order to prove this ,

Let us use the formula :

x3 + y3 + z3 - 3xyz = ( x + y + z) (x2 + y2 + z2 - xy - yz - xz)

Now let us keep x + y + z = 0

So,

x3 + y3 + z3 - 3xyz = 0

x3 + y3 + z3 = 3xyz

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# I hope this will help you !!!!