Question

If $a,b,c$ are three integers. All three of them cannot have the same value (any two of them can be equal). $w=\sqrt[3]{31},(w\ne 1)$, then the minimum values of $|a+bw+c{w}^2|$

• A. $0$
• B. $1$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$1$

Since $w$ is the cube root of unity. $\therefore w^3=1\&1+w+w^2=0$, where $w=\frac{-1+i\sqrt{3}}{2}$ and $w^2=\frac{-1-i\sqrt{3}}{2}$.
${|a+bw+c{w}^2|}^2=(a+bw+c{w}^2)(a+b\overline{w}+c\overline{w^2})=a^2+b^2+c^2+(bc+ca+ab)(w^2+w)$
$\Rightarrow {|a+bw+c{w}^2|}^2=a^2+b^2+c^2-(bc+ca+ab)=\frac{{(a-b)}^2+{(b-c)}^2{+(c-a)}^2}{2}$
Since a,b,c are integers, not all equal, atleast two of $(a-b)$, $(b-c)$ and $(c-a)$
$\therefore {(a-b)}^2+{(b-c)}^2{+(c-a)}^2\ge 2\Rightarrow {|a+bw+c{w}^2|}^2\ge 1$ (Equality will occur when two numbers are equal and the third number differs by 1)
Hence, option B is correct.