Question

If $x=a+b,y=a\omega +b{\omega }^2$ and $z=a{\omega }^2+b\omega$, then xyz is equal to

• A. $(a+b{)}^3$
• B. $a^3+b^3$
• C. $a^3-b^3$
• D. $(a+b{)}^2-3ab(a+b)$

Answer 1

Answer: (B)

$a^3+b^3$

$1+\omega +{\omega }^2=0$

${\omega }^3=1$

$x=a+b$

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

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

$x+y+z=a(1+\omega +{\omega }^2)+b(1+\omega +{\omega }^2)$

As $x+y+z=0$ then

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

$x^3+y^3+z^3={(a+b)}^3+{(a\omega +b{\omega }^2)}^3+{(b\omega +a{\omega }^2)}^3$

$x^3+y^3+z^3=a^3(1+{\omega }^3+{\omega }^5)+b^3(1+{\omega }^3+{\omega }^5)$

$x^3+y^3+z^3=3(a^3+b^3)$

$3xyz=3(a^3+b^3)$

$xyz=(a^3+b^3)$

Thus Option B