Question

If a, b, c, are distinct integers and $w\ne 1$ is a cube root of unity, then minimum value of $x=|a+bw+c{w}^2|+|a+b{w}^2+cw|$

• A. $2$
• B. $3$
• C. $4\sqrt{2}$
• D. $6\sqrt{2}$

Answer 1

The correct option is D $6\sqrt{2}$

Let $z=z+bw+c{w}^2$ $\therefore \overline{z}=a+b{w}^2+cw$
$\therefore \left|z\right|=\left|\overline{z}\right|\cdots \left(1\right)$ and $z\overline{z}=a^2+b^2+c^2-bc-ca-ab$
${\left|z\right|}^2=\frac{1}{2}[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$
$\therefore x=\left|z\right|+{\left|\overline{z}\right|}^2=4.\frac{1}{2}\left[{(\sum (a-b))}^2\right]$
$\therefore x=\sqrt{2}[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]\cdots \left(2\right)$
Since, a, b, c are integers, x will be minimum if a, b, c are consecutive integers p, p +1,p+2
$\therefore x=\sqrt{2}[1+1+4]=6\sqrt{2}\Rightarrow \left(d\right).$