Question

If $\omega \ne 1$ is a cube root of unity, then the value of $(x+y{)}^2+(x\omega +y{\omega }^2{)}^2+(x{\omega }^2+y\omega {)}^2$ is

• A. $3xy$
• B. $6xy$
• C. $4xy$
• D. $xy$

Answer 1

The correct option is B $6xy$

Since $w$ is the cube root of unity.
$\therefore w^3=1\&1+w+w^2=0$
Let $z=(x+y{)}^2+(xw+y{w}^2{)}^2+(x{w}^2+yw{)}^2$
$\Rightarrow z=[x^2+y^2+2xy]+[{\left(xw\right)}^2+{\left(y{w}^2\right)}^2+2\left(xw\right)\left(y{w}^2\right)]+[{\left(x{w}^2\right)}^2+{\left(yw\right)}^2+2\left(x{w}^2\right)\left(yw\right)]$
$\Rightarrow z=x^2(1+w^2+w)+y^2(1+w+w^2)+2xy(1+w^3+w^3)$
$\Rightarrow z=6xy$
Ans: B