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. ${{p}^{3}}-{{q}^{3}}$

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.

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.