Understanding the Problem

The question asks us to find the value of the expression $x^2 + y^2 + z^2$, given the definitions of $x$, $y$, and $z$ in terms of $a$, $b$, and the cube roots of unity ($1, \omega, \omega^2$).

Key Properties of Cube Roots of Unity

Before we start calculating, let's recall the essential properties of the cube roots of unity:

• The cube roots of unity are $1$, $\omega$, and $\omega^2$.
• They satisfy the equation $t^3 = 1$.
• The sum of the cube roots of unity is zero: $1 + \omega + \omega^2 = 0$.
• Also, $\omega^3 = 1$.
• From $1 + \omega + \omega^2 = 0$, we can derive other relations like $1 + \omega = -\omega^2$, $1 + \omega^2 = -\omega$, and $\omega + \omega^2 = -1$.
• Also $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$.

Given Expressions

We are given:

• $x = a + b$
• $y = a\omega + b\omega^2$
• $z = a\omega^2 + b\omega$

Calculating the Squares

Now, let's calculate the square of each expression:

1. Calculating $x^2$:

   Using the formula $(p+q)^2 = p^2 + 2pq + q^2$: $x^2 = (a + b)^2 = a^2 + 2ab + b^2$

2. Calculating $y^2$:

   Using the same formula: $y^2 = (a\omega + b\omega^2)^2$ $y^2 = (a\omega)^2 + 2(a\omega)(b\omega^2) + (b\omega^2)^2$ $y^2 = a^2\omega^2 + 2ab\omega^3 + b^2\omega^4$ Using the properties $\omega^3 = 1$ and $\omega^4 = \omega$: $y^2 = a^2\omega^2 + 2ab(1) + b^2\omega$ $y^2 = a^2\omega^2 + 2ab + b^2\omega$

3. Calculating $z^2$:

   Similarly: $z^2 = (a\omega^2 + b\omega)^2$ $z^2 = (a\omega^2)^2 + 2(a\omega^2)(b\omega) + (b\omega)^2$ $z^2 = a^2\omega^4 + 2ab\omega^3 + b^2\omega^2$ Using the properties $\omega^3 = 1$ and $\omega^4 = \omega$: $z^2 = a^2\omega + 2ab(1) + b^2\omega^2$ $z^2 = a^2\omega + 2ab + b^2\omega^2$

Summing the Squares

Now, we need to find the sum $x^2 + y^2 + z^2$: $x^2 + y^2 + z^2 = (a^2 + 2ab + b^2) + (a^2\omega^2 + 2ab + b^2\omega) + (a^2\omega + 2ab + b^2\omega^2)$

Let's group the terms involving $a^2$, $ab$, and $b^2$: $x^2 + y^2 + z^2 = (a^2 + a^2\omega^2 + a^2\omega) + (2ab + 2ab + 2ab) + (b^2 + b^2\omega + b^2\omega^2)$

Factor out $a^2$, $2ab$, and $b^2$: $x^2 + y^2 + z^2 = a^2(1 + \omega^2 + \omega) + 2ab(1 + 1 + 1) + b^2(1 + \omega + \omega^2)$

Applying the Properties

Using the property $1 + \omega + \omega^2 = 0$: $x^2 + y^2 + z^2 = a^2(0) + 2ab(3) + b^2(0)$

Simplifying the expression: $x^2 + y^2 + z^2 = 0 + 6ab + 0$ $x^2 + y^2 + z^2 = 6ab$

Conclusion

Therefore, the value of $x^2 + y^2 + z^2$ is $6ab$. This matches option 1.