Question

If $1,\omega ,{\omega }^2$ are the cube roots of unity, then $(x+y+z)(x+y\omega +z{\omega }^2)(x+y{\omega }^2+z\omega )$ 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

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

$(x+y+z)(x+y+z{\omega }^2)(x+y{\omega }^2+z\omega )Using1+\omega +{\omega }^2=0=(x+y+z)(x+y(-1-{\omega }^2)+z{\omega }^2)(x+y(-1-\omega )+z\omega )=(x+y+z)\left[\right(x-y)-{\omega }^2(y-z\left)\right]\left[\right(x-y)-\omega (y-z\left)\right]=(x+y+z)\left[\right(x-y{)}^2+(y-z{)}^2+(x-y\left)\right(y-z\left)\right][using{\omega }^3=1]=(x+y+z)[x^2+y^2-2xy+y^2+z^2-2yz+xy-xz-y^2+yz]=(x+y+z)[x^2+y^2+z^2-xy-yz-xz]=x^3+y^3+z^3–3xyz$