Question

If $'\omega '$ is cube root of unity and $x+y+z=a$,$x+\omega y+{{\omega }^{2}}z=b$ and $x+{{\omega }^{2}}y+\omega z=c$ then which of the following is correct
( a ) $x=\dfrac{a+b+c}{3}$
( b )$y=\dfrac{a+b{{\omega }^{2}}+c\omega }{3}$
( c ) $z=\dfrac{a+c{{\omega }^{2}}+b\omega }{3}$
( d ) All of the above

Answer 1

Here we will use two cube root identities which are $1+\omega +{{\omega }^{2}}=0$ and ${{\omega }^{3}}=1$to solve the equations. Basically, in this question we will substitute values of x, y and z to find the answer matching any of the correct options.

Complete step by step answer:
We know that $1+\omega +{{\omega }^{2}}=0$ and ${{\omega }^{3}}=1$ , where $'\omega '$ is cube root of unity.
It is given that, $x+y+z=a,$…… ( i )
$x+\omega y+{{\omega }^{2}}z=b$ , …….( ii )
$x+{{\omega }^{2}}y+\omega z=c$………. ( iii )
Adding, equation ( i ), ( ii ), ( iii )
$x+y+z+x+\omega y+{{\omega }^{2}}z+x+{{\omega }^{2}}y+\omega z=a+b+c$
Taking common factors out together, we get
$3x+y\left( 1+\omega +{{\omega }^{2}} \right)+z\left( 1+\omega +{{\omega }^{2}} \right)=a+b+c$ …… ( iv )
As, we mentioned in above statement that $1+\omega +{{\omega }^{2}}=0$,
So substituting value of $1+\omega +{{\omega }^{2}}=0$in equation ( iv ), we get
3x = a + b + c,
Putting 3 from numerator of left hand side to denominator of right hand side, we get
$x=\dfrac{a+b+c}{3}$
Now it is given that, $x+y+z=a,$
So, we can re – write $x+y+z=a,$as $y+z=a-x$ …… ( v )
Putting value of x in equation ( v )
$y+z=a-\dfrac{(a+b+c)}{3}$
On solving, we get
$y+z=\dfrac{2a-b-c}{3}$
Shifting z from left hand side to right hand side, we get
$y=\dfrac{2a-b-c}{3}-z$
On solving, we get
$y=\dfrac{2a-b-c-3z}{3}$
Putting values of x and y in equation ( i ), we get
$\dfrac{a+b+c}{3}+\omega \left( \dfrac{2a-b-c-3z}{3} \right)+{{\omega }^{2}}z=b$…… ( vi )
Taking L.C.M in equation ( vi ), we get
$a+b+c+2a\omega -b\omega -c\omega -3z\omega +3{{\omega }^{2}}z=3b$
Taking 3b from right hand to left hand side, we get
$a-2b+c+2a\omega -b\omega -c\omega -3z\omega +3{{\omega }^{2}}z=0$
Moving $-3z\omega +3{{w}^{2}}z$ from left hand side to right hand side, we get
$a-2b+c+2a\omega -b\omega -c\omega =3z\omega -3{{\omega }^{2}}z$
On simplifying we get,
$\begin{aligned} & a(1+\omega )+a\omega -b(1+\omega )-b+c(1-\omega )=3z\omega \left( 1-\omega \right) \\\ & -a{{\omega }^{2}}+a\omega +b{{\omega }^{2}}-b+c(1-\omega )=3z\omega (1-\omega ) \\\ & a\omega (1-\omega )-b(1-\omega )(1+\omega )+c(1-\omega )=3z\omega (1-\omega ) \\\ & a\omega +b{{\omega }^{2}}+c=3z\omega \\\ \end{aligned}$
Putting $3\omega$ in denominator of left hand side, we get
$\dfrac{a\omega +b{{\omega }^{2}}+c}{3\omega }=z$
Multiplying, numerator and denominator of right hand side by ${{\omega }^{2}}$, we get
$z=\dfrac{(a\omega +b{{\omega }^{2}}+c)\cdot {{\omega }^{2}}}{3{{\omega }^{3}}}$
$z=\dfrac{a+b\omega +c{{\omega }^{2}}}{3}$……. ( vi )
Putting equation ( vi ) in equation $y=\dfrac{2a-b-c-3z}{3}$, we get
$y=\dfrac{2a-b-c-3\left( \dfrac{a+b\omega +c{{\omega }^{2}}}{3} \right)}{3}$
$y=\dfrac{2a-b-c-a-b\omega -c{{\omega }^{2}}}{3}$
$y=\dfrac{a-b(1+\omega )-c(1+{{\omega }^{2}})}{3}$,
As $-(1+\omega )={{\omega }^{2}}$ and $-(1+{{\omega }^{2}})=\omega$ , so putting these values in $y=\dfrac{a-b(1+\omega )-c(1+{{\omega }^{2}})}{3}$, we get
$y=\dfrac{a+b{{\omega }^{2}}+c\omega }{3}$
Option ( a ) is not correct as $'\omega '$ is cube root of unity and can take values 1, $-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ and $-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$, so option ( a ) will be correct only if it is mentioned that cube root of unity takes only real value.

So, the correct answer is “Option B”.

Note: The two general identities $1+\omega +{{\omega }^{2}}=0$ and ${{\omega }^{3}}=1$ should always be kept in mind while solving questions based on cube root unity. Also, we have to keep in mind while solving these questions, that we have to break down complex equations into simplest form so as to obtain a solution. Sign scheme should be placed properly else it may change the answer.