Questions & Answers

Question

Answers

$\begin{align}

& \text{A}\text{. }\left( x-y \right)\left( x-\omega y \right)\left( x-{{\omega }^{2}}y \right) \\

& \text{B}\text{. }\left( x-y \right)\left( x+\omega y \right)\left( x+{{\omega }^{2}}y \right) \\

& \text{C}\text{. }\left( x+y \right)\left( x+\omega y \right)\left( x+{{\omega }^{2}}y \right) \\

& \text{D}\text{. }\left( x+y \right)\left( x-\omega y \right)\left( x-{{\omega }^{2}}y \right) \\

\end{align}$

Answer
Verified

The cube of an imaginary cube root of unity is equal to one i.e. ${{\omega }^{3}}=1$

The sum of cube roots of unity is equal to zero i. e. $1+\omega +{{\omega }^{2}}=0$

We have given that $\omega $ is the cube root of unity.

We have to find the linear factors of ${{x}^{3}}+{{y}^{3}}$ in complex numbers.

Now, we know that the cube root of unity is that number which when raised to power $3$ gives the answer as $1$ . There are three values of cube root of unity from which two are complex cube roots of unity and one is real cube root.

The three cube roots of unity are $1,\left( -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right),\left( -\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \right)$ also represented as $1,\omega ,{{\omega }^{2}}$.

Now, we have given a linear equation ${{x}^{3}}+{{y}^{3}}$.

We know that ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$

So, the given equation becomes

$\Rightarrow {{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$

We can write the equation as ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-1.xy+1.{{y}^{2}} \right)$

Now, we know that The sum of cube roots of unity is equal to zero i. e. $1+\omega +{{\omega }^{2}}=0$

Or $1=-\left( \omega +{{\omega }^{2}} \right)$

The cube of an imaginary cube root of unity is equal to one i.e. ${{\omega }^{3}}=1$

Now, the above equation will be ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}+\left( \omega +{{\omega }^{2}} \right)xy+{{\omega }^{3}}{{y}^{2}} \right)$

When we simplify further, we get

\[\begin{align}

& {{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}+\omega xy+{{\omega }^{2}}xy+{{\omega }^{3}}{{y}^{2}} \right) \\

& {{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( x+\omega y \right)\left( x+{{\omega }^{2}}y \right) \\

\end{align}\]

So, the linear factors of ${{x}^{3}}+{{y}^{3}}$ in complex number is \[\left( x+y \right)\left( x+\omega y \right)\left( x+{{\omega }^{2}}y \right)\]