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

& =xyz \\\ & =\left( p+q \right)\left( p\omega +q{{\omega }^{2}} \right)\left( p{{\omega }^{2}}+q\omega \right) \\\ \end{aligned}$$ Multiplying each value inside brackets, we get $$\begin{aligned} & =\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{aligned}$$ Now, rearranging the similar terms together in the above equation, we get $$\begin{aligned} & =\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{aligned}$$ Taking out the common terms from each of the brackets, we get $$\begin{aligned} & =\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{aligned}$$ Now, we need to remove $\omega $ and for that, we have to use the properties of the complex cube root of unity, i.e., $1+\omega +{{\omega }^{2}}=0$ and ${{\omega }^{3}}=1$ Also, $${{\omega }^{4}}={{\omega }^{3}}\cdot \omega =1\cdot \omega =\omega $$ Thus, substituting these values in equation (1), we get $$\begin{aligned} & =\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{aligned}$$ And, $\begin{aligned} & \Rightarrow 1+\omega +{{\omega }^{2}}=0 \\\ & \Rightarrow \omega +{{\omega }^{2}}=-1 \\\ \end{aligned}$ Thus, substituting this as well, we get $$\begin{aligned} & =\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{aligned}$$ By multiplying these terms, we finally get $$\begin{aligned} & =\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{aligned}$$ 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.