To show that if [tex]\( x + y + z = 0 \)[/tex], then [tex]\( x^3 + y^3 + z^3 = 3xyz \)[/tex], let's proceed with the following steps:
### Step 1: Use the Identity for the Sum of Cubes
We start with a well-known algebraic identity for the sum of cubes:
[tex]\[
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
\][/tex]
### Step 2: Substitute the Given Condition
Given that [tex]\( x + y + z = 0 \)[/tex], we substitute this into the identity:
[tex]\[
x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)
\][/tex]
### Step 3: Simplify the Equation
Since any number multiplied by 0 is 0, the right-hand side of the equation becomes 0:
[tex]\[
x^3 + y^3 + z^3 - 3xyz = 0
\][/tex]
### Step 4: Rearrange to Show the Desired Result
Rearranging the equation, we get:
[tex]\[
x^3 + y^3 + z^3 = 3xyz
\][/tex]
### Verification with Example Values
To confirm this identity, let's take specific values of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] that satisfy [tex]\( x + y + z = 0 \)[/tex]. For instance, let [tex]\( x = 1 \)[/tex], [tex]\( y = -2 \)[/tex], and [tex]\( z = 1 \)[/tex]:
[tex]\[
x + y + z = 1 - 2 + 1 = 0
\][/tex]
This satisfies the given condition. Now, we calculate both sides of the equation:
1. Calculate the left side:
[tex]\[
x^3 + y^3 + z^3 = 1^3 + (-2)^3 + 1^3 = 1 - 8 + 1 = -6
\][/tex]
2. Calculate the right side:
[tex]\[
3xyz = 3 \cdot 1 \cdot (-2) \cdot 1 = 3 \cdot (-2) = -6
\][/tex]
Both sides of the equation equal [tex]\(-6\)[/tex], confirming that the relation [tex]\( x^3 + y^3 + z^3 = 3xyz \)[/tex] holds true for these values.
Thus, we have shown that if [tex]\( x + y + z = 0 \)[/tex], then [tex]\( x^3 + y^3 + z^3 = 3xyz \)[/tex].
