Question

If $\omega $ is cube root of unity the linear factors of ${{x}^{3}}+{{y}^{3}}$ in complex number is
$\begin{align}
  & \text{A}\text{. }\left( x-y \right)\left( x-\omega y \right)\left( x-{{\omega }^{2}}y \right) \\
 & \text{B}\text{. }\left( x-y \right)\left( x+\omega y \right)\left( x+{{\omega }^{2}}y \right) \\
 & \text{C}\text{. }\left( x+y \right)\left( x+\omega y \right)\left( x+{{\omega }^{2}}y \right) \\
 & \text{D}\text{. }\left( x+y \right)\left( x-\omega y \right)\left( x-{{\omega }^{2}}y \right) \\
\end{align}$

Answer 1

Hint: To solve this question first we should know the basic concepts of complex numbers and cube root of unity. Start solving this question by using the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ to factorize the given equation and then use the properties of cube root of unity to solve further. Following properties we used to solve this question: If $\omega $ is cube root of unity then
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. $1+\omega +{{\omega }^{2}}=0$

Complete step by step answer:
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 \[\left( x+y \right)\left( x+\omega y \right)\left( x+{{\omega }^{2}}y \right)\]

So, the correct answer is “Option C”.

Note: It is important to know the basics of the cube root of unity and complex numbers to solve such types of questions. Cube roots of unity are found using the factoring equations and solving quadratic equations. Complex numbers are expressed in the form $\left( a+ib \right)$, where $a\And b$ are real numbers and $i$ is the imaginary number.