Question

If $1,\omega ,w^2$ are the cube roots of unity then the value of $(x+y+z)(x+y\omega +z{\omega }^2)(x+y{\omega }^2+z\omega )$is equal to?

• A. x + y + z
• B. $x^2+y^2+z^2$
• C. $x^3+y^3+z^3$
• D. $x^3+y^3+z^3-3xyz$

Answer 1

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

Expanding the given expression step by step,
$(x+y+z)(x+y\omega +3{\omega }^2)(x+y{\omega }^2+z\omega )$
$=(x^2+xy\omega +\underset{–}{3x}{\omega }^2+2y+y^2\omega +yz{\omega }^2+\underset{–}{xz}+yz\omega +z^2{\omega }^2)\times (x+y{\omega }^2+z\omega )$
$=[x^2+y^2\omega +z^2{\omega }^2+xy(1+\omega )+xz(1+{\omega }^2)+zy(\omega +{\omega }^2\left)\right][x+y{\omega }^2+z\omega ]$
$=x^3+y^3{\omega }^3+z^3{\omega }^3+x\left[xy\right(1+\omega )+xz(1+{\omega }^2)+yz({\omega }^{-1}+{\omega }^2\left)\right]$
$+y{\omega }^2\left[xy\right(1+\omega )+xz(1+{\omega }^2)+yz({\omega }^{-1}+{\omega }^2\left)\right]$
$+z\omega \left[xy\right(1+\omega )+xz(1+{\omega }^2)+yz({\omega }^{-1}+{\omega }^2\left)\right]$
$+x{y}^2\omega +x{z}^2{\omega }^2+x^{2}y{\omega }^2+y{z}^2\omega +x^{2}z\omega +y^{2}z{\omega }^2$
$=x^3+y^3+z^3+x^{2}y(1+\omega +{\omega }^2)+x^{2}z(1+\omega +{\omega }^2)-xyz+x{y}^2(1+\omega +{\omega }^2)$
$+z{y}^2(1+\omega +{\omega }^2)-xyz+x{z}^2(1+\omega +{\omega }^2)+y{z}^2(1+\omega +{\omega }^2)-xyz$
$=x^3+y^3+z^3-3xyz$ ------------$\because (1+\omega +{\omega }^2=0)$