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.