Let , , be three cube roots of unity. If , , , then what is equal to?

6ab. Squaring each term gives , , and . Since , the sum is . Using the identity , the and terms vanish, leaving .

Let 1, \omega, \omega^2 be three cube roots of unity. If x=a+b, y=a\omega+b\omega^2, z=a\omega^2+b\omega, then what is x^2+y^2+z^2 equal to?

A. 6ab ✓
B. 3ab
C. a^2+b^2
D. 1

Correct Answer: A. 6ab

Explanation

Squaring each term gives x^2 = a^2+b^2+2ab, y^2 = a^2\omega^2+b^2\omega+2ab\omega^3, and z^2 = a^2\omega+b^2\omega^2+2ab\omega^3. Since \omega^3=1, the sum is x^2+y^2+z^2 = a^2(1+\omega+\omega^2) + b^2(1+\omega+\omega^2) + 2ab(1+1+1). Using the identity 1+\omega+\omega^2=0, the a^2 and b^2 terms vanish, leaving 6ab.

Let 1, \omega, \omega^2 be three cube roots of unity. If x=a+b, y=a\omega+b\omega^2, z=a\omega^2+b\omega, then what is x^2+y^2+z^2 equal to?

Explanation

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