Question

$\alpha ,\beta$ are complex cube roots of unity and $x=a+b,y=a\alpha +b\beta$ , $z=a\beta +b\alpha$ then $x^3+y^3+z^3=$

• A. $0$
• B. $3ab$
• C. $3(a^3+b^3)$
• D. $3(a+b{)}^3$

Answer 1

The correct option is C $3(a^3+b^3)$

$\alpha =\omega ,\beta ={\omega }^2$ ....... $[\because 1,\omega ,{\omega }^2$ are cube roots of unity]

$x=a+b$

$y=a\omega +b{\omega }^2$

$z=a{\omega }^2+b\omega$

$x+y+z=(a+b)+(a\omega +b{\omega }^2)+(a{\omega }^2+b\omega )$

$=a(1+\omega +{\omega }^2)+b(1+\omega +{\omega }^2)=0$ ...... $[\because {\omega }^3=1]$

Hence, $x^3+y^3+z^3=3xyz$

So, $xyz=(a+b)(a\omega +b{\omega }^2)(a{\omega }^2+b\omega )=3(a^3+b^3)$