Question

If $\alpha ,\beta ,\gamma$ are the cube roots of $p,p<0$, then for any $x,y$ & $z$ then

$\frac{x\alpha +y\beta +z\gamma }{x\beta +y\gamma +z\alpha }=$

• A. $\omega$
• B. ${\omega }^3$
• C. ${\omega }^2$
• D. 1

Answer 1

The correct option is C ${\omega }^2$

If $x^3=1$
$1,w,w^{2}arethecuberootsofunity.$
$⟹1+w+w^2=0$
where, $w=\frac{-1+i\sqrt{3}}{2};w^2=\frac{-1-i\sqrt{3}}{2}$
Let z be the cube root of p.
${⟹z}^3=p$
$⟹z=p^{\frac{1}{3}}{\left(cis\left(0\right)\right)}^{\frac{1}{3}}$ ...{De Moivre's Theorem}
$=p^{\frac{1}{3}}cis\left(\frac{2k\Pi }{3}\right)$
where, k=0,1,2.
$\alpha =p^{\frac{1}{3}}\beta =p^{\frac{1}{3}}cis\left(\frac{2\Pi }{3}\right)=p^{\frac{1}{3}}w\gamma =p^{\frac{1}{3}}cis\left(\frac{4\Pi }{3}\right)=p^{\frac{1}{3}}{w}^2$
$\frac{x\alpha +y\beta +z\gamma }{x\beta +y\gamma +z\alpha }=\frac{x+yw+z{w}^2}{xw+y{w}^2+z}=\frac{w^2(x+yw+z{w}^2)}{x{w}^3+y{w}^4+z{w}^2}=w^2$
Ans: C