Question

$(x+y+z)(x+y\omega +z{\omega }^2)(x+y{\omega }^2+z\omega )=E$, where $\omega$ is the cube root of unity. Then the value of $E$ is

• A. $x^3+y^3+z^3+3xyz$
• B. $x^3+y^3+z^3-3xyz$
• C. $x^3+y^3+z^3$
• D. $3xyz$

Answer 1

The correct option is B $x^3+y^3+z^3-3xyz$

$E=(x+y+z)(x+y\omega +z{\omega }^2)(x+y{\omega }^2+z\omega )$
$\because {\omega }^3=1$
$\Rightarrow E=(x+y+z)(x^2+xy{\omega }^2+xz\omega +xy\omega +y^2+yz{\omega }^2+xz{\omega }^2+yz\omega +z^2)$
$\Rightarrow E=(x+y+z)(x^2+y^2+z^2+xy(\omega +{\omega }^2)+yz(\omega +{\omega }^2)+xz(\omega +{\omega }^2\left)\right)$
$\because 1+\omega +{\omega }^2=0$
$\Rightarrow E=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
$\Rightarrow E=(x^3+y^3+z^3-3xyz)$