lf the roots of the cubic equation ${(z + αβ)}^{3} = α^{3}, α \neq 0$ represent the vertices of a triangle of sides of length _________

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$\frac{1}{\sqrt{3}} |αβ|$

$\sqrt{3} |α|$

$\frac{1}{\sqrt{3}} |α|$

$\sqrt{3} |β|$

answer is B.

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Detailed Solution

$Given {(z + αβ)}^{3} = α^{3} \Rightarrow z + αβ = {(α^{3} . 1)}^{\frac{1}{3}}$

$\Rightarrow z + αβ = α {(1)}^{\frac{1}{3}}$

$1, ω, ω^{2} a r e c u b r o o t s o f u n i t y$

$let z_{1}, z_{2}, z_{3} are vertices of triangle$

$Now z_{1} = α - αβ, z_{2} = α ω - α β, a n d z_{3} = α ω^{2} - α β$

$N o w |z_{1} - z_{2}| = |α| |1 - ω|$

$= |α| |1 - \frac{(- 1 + \sqrt{3} i)}{2}|$

$= |α| |\frac{3 - i \sqrt{3}}{2}|$

$= |α| \frac{\sqrt{12}}{2}$

$= \sqrt{3} |α|$

$N o w |z_{2} - z_{3}| = |α| |ω - ω^{2}|$

$= |α| |\frac{- 1 + i \sqrt{3}}{2} - (\frac{- 1 - i \sqrt{3}}{2})|$

$= \sqrt{3} |α|$

$a n d |z_{3} - z_{1}| = |α| |1 - ω^{2}| = \sqrt{3} |α|$

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