Question

If \(1, \omega, \omega^2\) are the cube roots of unity, then the value of \[ (x+y)^2 + (x\omega + y\omega^2)^2 + (x\omega^2 + y\omega)^2 \] is:

• A. \(2x^2 \cdot 3y^2\)
• B. \(4xy\)
• C. \(6xy\)
• D. \(2x^2 \cdot 2y^2\)

Hint:

When dealing with cube roots of unity, always use \(1 + \omega + \omega^2 = 0\) to simplify sums of powers.

Answer 1

The Correct Option is C

Step 1: Recall properties of cube roots of unity.
\[ 1 + \omega + \omega^2 = 0, \quad \omega^3 = 1 \]

Step 2: Expand each square term.
\[ (x+y)^2 = x^2 + y^2 + 2xy \] \[ (x\omega + y\omega^2)^2 = x^2 \omega^2 + y^2 (\omega^2)^2 + 2xy \omega \omega^2 = x^2 \omega^2 + y^2 \omega + 2xy \] \[ (x\omega^2 + y\omega)^2 = x^2 \omega + y^2 \omega^2 + 2xy \]

Step 3: Sum all three expressions.
\[ (x+y)^2 + (x\omega + y\omega^2)^2 + (x\omega^2 + y\omega)^2 \] \[ = (x^2 + x^2\omega^2 + x^2\omega) + (y^2 + y^2\omega + y^2\omega^2) + (2xy + 2xy + 2xy) \]

Step 4: Simplify powers using \(1 + \omega + \omega^2 = 0\).
\[ x^2 (1 + \omega + \omega^2) = 0, \quad y^2 (1 + \omega + \omega^2) = 0 \] \[ 2xy + 2xy + 2xy = 6xy \]

Step 5: Final result.
\[ \boxed{6xy} \]