# 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

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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.

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.

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