Question

If $x=a+b$, $y=a\omega +b{\omega }^2$, $z=a{\omega }^2+b\omega$, $\omega$ is cube root of unity then value of $x^3+y^3+z^3$

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

Answer 1

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

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

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

$=a^3+b^3+3a^{2}b+3b^{2}a+a^3{\omega }^3+b^3{\omega }^6+3a^{2}b{\omega }^4+3a{b}^2{\omega }^5+a^3{\omega }^6+b^3{\omega }^3+3a^{2}b{\omega }^5+3a{b}^2{\omega }^4$

$=a^3+b^3+3a^{2}b+3a{b}^2+a^3+b^3+3a^{2}b\omega +3a{b}^2{\omega }^2+a^3+b^3+3a^{2}b{\omega }^2+3a{b}^2\omega$

$=3(a^3+b^3)+3a^{2}b(1+\omega +{\omega }^2)+3a{b}^2(1+\omega +{\omega }^2)$

We know that $1+\omega +{\omega }^2=0$

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