Question

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

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

Answer 1

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

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