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: Name the three terms and note they sum to zero.
Let \(a=x+y\), \(b=x\omega+y\omega^2\), \(c=x\omega^2+y\omega\).
\(a+b+c = x(1+\omega+\omega^2)+y(1+\omega+\omega^2) = 0\), using \(1+\omega+\omega^2=0\).

Step 2: Use the identity relating the sum of squares to the sum and the pairwise products.
\(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)\), and since \(a+b+c=0\), this becomes \(a^2+b^2+c^2=-2(ab+bc+ca)\).

Step 3: Compute ab+bc+ca.
Expanding each product and repeatedly using \(\omega^3=1\) and \(1+\omega+\omega^2=0\), the \(x^2\) and \(y^2\) terms cancel out and only the cross terms survive, giving \(ab+bc+ca=-3xy\).

Step 4: Conclusion.
\[ a^2+b^2+c^2=-2(-3xy)=6xy \]
\[ \boxed{6xy} \]