Question

If $\omega$ is the complex cube root of unity, then $(3 + 5\omega + 3\omega^2)^2 + (3 + 3\omega + 5\omega^2)^2 =$

• A. $-1$
• B. 0
• C. 4
• D. $-4$

Hint:

Whenever you see a quadratic trigonometric or polynomial structure with symmetric coefficients like $3 + 5\omega + 3\omega^2$, match the outer numbers first. Changing the coefficients of the outer terms to match the middle term using combinations of $1+\omega+\omega^2=0$ is almost always the fastest way to reduce the expression!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Simplify (3+5ω+3ω²)² + (3+3ω+5ω²)² using cube root of unity properties: 1+ω+ω²=0 and ω³=1.

Step 2: Key Formula or Approach:
Group terms to exploit 1+ω² = –ω and 1+ω = –ω² to simplify each bracket before squaring.

Step 3: Detailed Explanation:
First bracket: 3+5ω+3ω² = 3(1+ω²)+5ω = –3ω+5ω = 2ω. Second bracket: 3+3ω+5ω² = 3(1+ω)+5ω² = –3ω²+5ω² = 2ω². Sum of squares = 4ω²+4ω⁴ = 4ω²+4ω = 4(ω²+ω) = 4(–1) = –4.

Step 4: Final Answer:
The value is –4, matching option (D).