Question

If $x=a+b,y=a\omega +\beta {\omega }^2$ and $z=a{\omega }^2+b\omega$, then $x^3+y^3+z^3=$

• A. $a^3+b^3$
• B. $3(a^3+b^3)$
• C. $a^3-b^3$
• D. $3(a^3-b^3)$

Answer 1

Answer: (B)

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

We have, $x+y+z=(a+b)+(a\omega +b{\omega }^2)+(a{\omega }^2+b\omega )$
$=a(1+\omega +{\omega }^2)+b(1+\omega +{\omega }^2)$
$=a\times 0+b\times 0=0$ -----------(1)
Hence from the relation
$x^3+y^3+y^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-zx-xy)=0by$ (1),
we obtain $x^3+y^3+z^3=3xyz$
but $xyz=a^3+b^3$
Hence $x^3+y^3+z^3=3(a^3+b^3)$