Q167 / 270 · NDA 2025 (Phase 2) · Mathematics
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?
- Correct answerA$6ab$
- B$3ab$
- C$a^2 + b^2$
- D$1$
Solution
Given $x = a + b$, $y = a\omega + b\omega^2$, and $z = a\omega^2 + b\omega$.
Squaring each term, we get:
$x^2 = (a + b)^2 = a^2 + b^2 + 2ab$
$y^2 = (a\omega + b\omega^2)^2 = a^2\omega^2 + b^2\omega^4 + 2ab\omega^3$
$z^2 = (a\omega^2 + b\omega)^2 = a^2\omega^4 + b^2\omega^2 + 2ab\omega^3$
Using the properties of cube roots of unity, $\omega^3 = 1$ and $\omega^4 = \omega$. Substituting these into the equations:
$y^2 = a^2\omega^2 + b^2\omega + 2ab$
$z^2 = a^2\omega + b^2\omega^2 + 2ab$
Adding $x^2$, $y^2$, and $z^2$:
$x^2 + y^2 + z^2 = a^2(1 + \omega + \omega^2) + b^2(1 + \omega + \omega^2) + 2ab(1 + 1 + 1)$
Since $1 + \omega + \omega^2 = 0$, the expression simplifies to:
$x^2 + y^2 + z^2 = a^2(0) + b^2(0) + 6ab = 6ab$
Squaring each term, we get:
$x^2 = (a + b)^2 = a^2 + b^2 + 2ab$
$y^2 = (a\omega + b\omega^2)^2 = a^2\omega^2 + b^2\omega^4 + 2ab\omega^3$
$z^2 = (a\omega^2 + b\omega)^2 = a^2\omega^4 + b^2\omega^2 + 2ab\omega^3$
Using the properties of cube roots of unity, $\omega^3 = 1$ and $\omega^4 = \omega$. Substituting these into the equations:
$y^2 = a^2\omega^2 + b^2\omega + 2ab$
$z^2 = a^2\omega + b^2\omega^2 + 2ab$
Adding $x^2$, $y^2$, and $z^2$:
$x^2 + y^2 + z^2 = a^2(1 + \omega + \omega^2) + b^2(1 + \omega + \omega^2) + 2ab(1 + 1 + 1)$
Since $1 + \omega + \omega^2 = 0$, the expression simplifies to:
$x^2 + y^2 + z^2 = a^2(0) + b^2(0) + 6ab = 6ab$
Hence, the correct answer is option: 'A'.