Question

If $\omega \ne 1$ is a cube root of unity and $x+y+z=a$, $x+\omega y+{\omega }^{2}z=b$, $x+{\omega }^{2}y+\omega z=c$, then

• A. $x=(a+b+c)/3$
• B. $y=(a+b{\omega }^2+c\omega )/3$
• C. $z=(a+b\omega +c{\omega }^2)/3$
• D. $x^3+y^3+z^3=a^3/3$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $x=(a+b+c)/3$
B $y=(a+b{\omega }^2+c\omega )/3$
C $z=(a+b\omega +c{\omega }^2)/3$
D $x^3+y^3+z^3=a^3/3$

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$
$\Delta =\left|\begin{array}{ccc}3& 1& 1\\ 0& \omega & {\omega }^2\\ 0& {\omega }^2& \omega \end{array}\right|=3({\omega }^2-{\omega }^4)=3({\omega }^2-\omega )$
Now
${\Delta }_1=\left|\begin{array}{ccc}a& 1& 1\\ b& \omega & {\omega }^2\\ c& {\omega }^2& \omega \end{array}\right|$
Applying $R_1\to R_1+R_2+R_3$
${\Delta }_1=\left|\begin{array}{ccc}a+b+c& 0& 0\\ b& \omega & {\omega }^2\\ c& {\omega }^2& \omega \end{array}\right|=(a+b+c)({\omega }^2-\omega )$
${\Delta }_2=\left|\begin{array}{ccc}1& a& 1\\ 1& b& {\omega }^2\\ 1& c& \omega \end{array}\right|$
Applying $R_1\to R_1+{\omega }^2{R}_2+\omega R_3$
${\Delta }_2=\left|\begin{array}{ccc}0& a+{\omega }^{2}b+\omega c& 0\\ 1& b& {\omega }^2\\ 1& c& \omega \end{array}\right|=(a+{\omega }^{2}b+\omega c)({\omega }^2-\omega )$
Similarly
${\Delta }_3=(a+\omega b+{\omega }^{2}c)({\omega }^2-\omega )$
Hence from cramer's rule
$x=\frac{{\Delta }_1}{\Delta }=\frac{a+b+c}{3},y=\frac{{\Delta }_2}{\Delta }=\frac{(a+{\omega }^{2}b+\omega c)}{3},z=\frac{{\Delta }_3}{\Delta }=\frac{(a+\omega b+{\omega }^{2}c)}{3}$