Question

If \[x=p+q,\text{ }y=p\omega +q{{\omega }^{2}}\] and $z=p{{\omega }^{2}}+q\omega $ where \[\omega \] is a complex cube root of unity, then $xyz$ =
A. ${{p}^{3}}+{{q}^{3}}$
B. \[{{p}^{2}}-pq+{{q}^{2}}\]
C. \[1+{{p}^{3}}+{{q}^{3}}\]
D. \[{{p}^{3}}-{{q}^{3}}\]

Answer 1

Hint: We can solve the given set of equations just by simply substituting the values of $x,\text{ }y$and $z$in $xyz$. Since the solution of $xyz$ is independent of \[\omega \], thus we have to keep that point in mind to use the property of the complex cube root of unity.

Complete step-by-step answer:
Here, we have $x=p+q,\text{ }y=p\omega +q{{\omega }^{2}}$ and $z=p{{\omega }^{2}}+q\omega $, where \[\omega \] is a complex cube root of unity.
And we have to find the value of $xyz$, thus by substituting the values of $x,\text{ }y$and $z$in it, we get
\[\begin{align}
  & =xyz \\
 & =\left( p+q \right)\left( p\omega +q{{\omega }^{2}} \right)\left( p{{\omega }^{2}}+q\omega \right) \\
\end{align}\]
Multiplying each value inside brackets, we get
 \[\begin{align}
  & =\left( p+q \right)\left( p\omega +q{{\omega }^{2}} \right)\left( p{{\omega }^{2}}+q\omega \right) \\
 & =\left( p+q \right)\left( p\omega \left( p{{\omega }^{2}}+q\omega \right)+q{{\omega }^{2}}\left( p{{\omega }^{2}}+q\omega \right) \right) \\
 & =\left( p+q \right)\left( {{p}^{2}}{{\omega }^{3}}+pq{{\omega }^{2}}+qp{{\omega }^{4}}+{{q}^{2}}{{\omega }^{3}} \right) \\
\end{align}\]
Now, rearranging the similar terms together in the above equation, we get
\[\begin{align}
  & =\left( p+q \right)\left( {{p}^{2}}{{\omega }^{3}}+pq{{\omega }^{2}}+qp{{\omega }^{4}}+{{q}^{2}}{{\omega }^{3}} \right) \\
 & =\left( p+q \right)\left( pq{{\omega }^{2}}+qp{{\omega }^{4}}+{{p}^{2}}{{\omega }^{3}}+{{q}^{2}}{{\omega }^{3}} \right) \\
\end{align}\]
Taking out the common terms from each of the brackets, we get
 \[\begin{align}
  & =\left( p+q \right)\left( pq{{\omega }^{2}}+qp{{\omega }^{4}}+{{p}^{2}}{{\omega }^{3}}+{{q}^{2}}{{\omega }^{3}} \right) \\
 & =\left( p+q \right)\left( pq\left( {{\omega }^{2}}+{{\omega }^{4}} \right)+{{\omega }^{3}}\left( {{p}^{2}}+{{q}^{2}} \right) \right)...\text{ }\left( 1 \right) \\
\end{align}\]
Now, we need to remove $\omega $ and for that, we have to use the properties of the complex cube root of unity, i.e.,
$1+\omega +{{\omega }^{2}}=0$ and ${{\omega }^{3}}=1$
Also, \[{{\omega }^{4}}={{\omega }^{3}}\cdot \omega =1\cdot \omega =\omega \]
Thus, substituting these values in equation (1), we get
\[\begin{align}
  & =\left( p+q \right)\left( pq\left( {{\omega }^{2}}+{{\omega }^{4}} \right)+{{\omega }^{3}}\left( {{p}^{2}}+{{q}^{2}} \right) \right) \\
 & =\left( p+q \right)\left( pq\left( {{\omega }^{2}}+\omega \right)+\left( 1 \right)\left( {{p}^{2}}+{{q}^{2}} \right) \right) \\
\end{align}\]
And,
 $\begin{align}
  & \Rightarrow 1+\omega +{{\omega }^{2}}=0 \\
 & \Rightarrow \omega +{{\omega }^{2}}=-1 \\
\end{align}$
Thus, substituting this as well, we get
\[\begin{align}
  & =\left( p+q \right)\left( pq\left( -1 \right)+\left( 1 \right)\left( {{p}^{2}}+{{q}^{2}} \right) \right) \\
 & =\left( p+q \right)\left( -pq+{{p}^{2}}+{{q}^{2}} \right) \\
 & =\left( p+q \right)\left( {{p}^{2}}+{{q}^{2}}-pq \right) \\
\end{align}\]
By multiplying these terms, we finally get
\[\begin{align}
  & =\left( p+q \right)\left( {{p}^{2}}+{{q}^{2}}-pq \right) \\
 & =p\left( {{p}^{2}}+{{q}^{2}}-pq \right)+q\left( {{p}^{2}}+{{q}^{2}}-pq \right) \\
 & ={{p}^{3}}+p{{q}^{2}}-{{p}^{2}}q+q{{p}^{2}}+{{q}^{3}}-p{{q}^{2}} \\
 & ={{p}^{3}}+{{q}^{3}} \\
\end{align}\]
Hence, $xyz={{p}^{3}}+{{q}^{3}}$.

Note: One thing to avoid here is direct multiplication of all the terms together after substituting the values of $x,\text{ }y\text{ and }z$, as that might lead to an error. Because of complex multiplications, it is advised to first multiply the terms consisting of $\omega $ only to eliminate confusion, by using properties of the complex cube root of unity.