Question

The value of the expression $\left( 1+\sqrt{\omega }+\omega \right)\left( 1-\sqrt{\omega }+\omega \right)=.$
(a)0
(b)1
(c)$\omega $
(d)${{\omega }^{2}}$

Answer 1

Hint: The cube roots of unity can be defined as the numbers which when raised to the power of 3 gives the result as 1 and is denoted by $\omega $. The three cube roots of unity are 1, $\omega $ and ${{\omega }^{2}}$. The sum of the three cube roots of unity is zero $\left( 1+\omega +{{\omega }^{2}}=0 \right)$.

Complete step-by-step answer:
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 mathematics like number theory. It is also called the De Moivre system.

Let us consider the given expression and rearranging the terms,

\[\left( 1+\sqrt{\omega }+\omega \right)\left( 1-\sqrt{\omega }+\omega \right)=.\left[ \left( 1+\omega \right)+\sqrt{\omega } \right]\left[ \left( 1+\omega \right)-\sqrt{\omega } \right]\]

Applying the algebraic formula $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ on the right side, we get
\[\left( 1+\sqrt{\omega }+\omega \right)\left( 1-\sqrt{\omega }+\omega \right)=.\left[ {{\left( 1+\omega \right)}^{2}}-{{\left( \sqrt{\omega } \right)}^{2}} \right]\]

Applying another algebraic formula ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ on the right side, we get

\[\left( 1+\sqrt{\omega }+\omega \right)\left( 1-\sqrt{\omega }+\omega \right)=.\left[ 1-2\omega +{{\omega }^{2}}-\omega \right]\]

Rearranging the term on right side, we get

\[\left( 1+\sqrt{\omega }+\omega \right)\left( 1-\sqrt{\omega }+\omega \right)=.\left[ {{\omega }^{2}}+\omega +1 \right]\]

We know that the sum of the three cube roots of unity is zero$\left( 1+\omega +{{\omega }^{2}}=0 \right)$.

\[\left( 1+\sqrt{\omega }+\omega \right)\left( 1-\sqrt{\omega }+\omega \right)=0\]

Hence the value of the given expression is zero.

Therefore, the correct option for the given question is option (a).

Note: You might get confused by the question, Is cube root of unity collinear? The answer is yes. Collinear means all three roots lie on the one line. As $1+\omega +{{\omega }^{2}}=0$, it can be said that the cube roots of unity are collinear.