Question

If $1,\omega ,{{\omega }^{2}}$ are three cube roots of unity, then $(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})$ is…

Answer 1

Hint: Here $1,\omega ,{{\omega }^{2}}$ are three cube roots of unity where ${{\omega }^{3}}=1$and$1+\omega +{{\omega }^{2}}=0$. First, simplify $(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})$ and then apply${{\omega }^{3}}=1$and$1+\omega +{{\omega }^{2}}=0$. Try it and you will get the answer.

Complete step-by-step answer:
A complex number is a number that can be expressed in the form $a+ib$, where$a$and $b$ are real numbers, and $i$ is a solution of the equation ${{x}^{2}}=-1$. Because no real number satisfies this equation, $i$ is called an imaginary number. For the complex number $a+ib$, $a$ is called the real part, and $b$ is called the imaginary part. Despite the historical nomenclature "imaginary", complex numbers are regarded in the mathematical sciences as just as "real" as the real numbers and are fundamental in many aspects of the scientific description of the natural world.
The root of unity is a number which is complex in nature and gives $1$ if raised to the power of a positive integer $n$. These roots are used in different branches and topics of maths like number theory. It is also called the de’Moivre system. Here we will discuss the cube roots of unity in detail.
The cube roots of unity can be defined as the numbers which when raised to the power of $3$ gives the result as $1$. In simple words, the cube root of unity is the cube root of $1$ i.e. ${{1}^{\dfrac{2}{3}}}$.
There are a total of three cube roots of unity which are as follows:
$1$which is real, $\dfrac{-1\pm i\sqrt{3}}{2}$which is complex.
As $1+\omega +{{\omega }^{2}}=0$ it can be said that the cube root of unity is collinear.
The cube roots of unity are $1,\omega ,{{\omega }^{2}}$. So, the product of cube roots of unity $=1\times \omega \times {{\omega }^{2}}={{\omega }^{3}}=1$.
Now it is given in the question that $1,\omega ,{{\omega }^{2}}$are three cube roots of unity.
We have to find $(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})$.
Now solving,$(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})=(1+\omega -{{\omega }^{2}}-\omega -{{\omega }^{2}}+{{\omega }^{3}}+{{\omega }^{2}}+{{\omega }^{3}}-{{\omega }^{4}})=(1-{{\omega }^{2}}+2{{\omega }^{3}}-{{\omega }^{4}})$
$(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})=(1-{{\omega }^{2}}+2{{\omega }^{3}}-{{\omega }^{4}})$
Now we know${{\omega }^{3}}=1$and$1+\omega +{{\omega }^{2}}=0$.
In the above equation, we have${{\omega }^{3}}=1$.
$(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})=(3-{{\omega }^{2}}-{{\omega }^{4}})$

$(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})=(3-{{\omega }^{2}}(1+{{\omega }^{2}}))$……… (1)
So above we can see$1+\omega +{{\omega }^{2}}=0$.
Now simplifying $1+\omega +{{\omega }^{2}}=0$ we get,
$1+{{\omega }^{2}}=-\omega$…… (2)
Substituting (2) in (1), we get,
$(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})=(3-{{\omega }^{2}}(-\omega ))=(3+{{\omega }^{3}})$
We know${{\omega }^{3}}=1$.
$(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})=(3+{{(1)}^{3}})=4$
Here finally we get the value, of $(1-\omega +{{\omega }^{2}})(1+\omega -{{\omega }^{2}})$ which is equal to $4$.

Note: Carefully read the question. Do not jumble yourself. You should be familiar with the concept of the cube root of unity. You should also know that ${{\omega }^{3}}=1$and $1+\omega +{{\omega }^{2}}=0$. While simplification please be careful. Do not miss any term.