Question

If omega ($\omega $) is an imaginary cube root of unity and $x = a + b$, $y = a\omega+ b{\omega ^2}$, $z = a{\omega ^2} + b\omega $, then ${x^2} + {y^2} + {z^2}$ is equal to
1). $6ab$
2). $3ab$
3). $6{a^2}{b^2}$
4). $3{a^2}{b^2}$

Answer 1

Hint: In order to find the value of ${x^2} + {y^2} + {z^2}$, substitute the value of the variables given that is ${x^2},{y^2},{z^2}$. Expand the square using the expansion formula of squares and then take out the common values from them. Solve it further using the properties of omega and get the results.
Formula used:
1. ${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$
2. \[\left( {1 + \omega + {\omega ^2}} \right) = 0\]
3. \[{\omega ^3} = 1\]

Complete step-by-step solution:
We are given with three equations, that are stated as:
$x = a + b$
$y = a\omega + b{\omega ^2}$
$z = a{\omega ^2} + b\omega $
We need to find the value of ${x^2} + {y^2} + {z^2}$.
So, substituting the value of $x$, $y$ and $z$ in the equation ${x^2} + {y^2} + {z^2}$, and we get:
${x^2} + {y^2} + {z^2} = {\left( {a + b} \right)^2} + {\left( {a\omega + b{\omega ^2}} \right)^2} + {\left( {a{\omega ^2} + b\omega } \right)^2}$ …….(1)
From the algebraic expansion formula, we know that
${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$
So, using this formula for expanding the brackets with square power in the equation 1, and we get:
\[ \Rightarrow {x^2} + {y^2} + {z^2} = \left( {{a^2} + {b^2} + 2ab} \right) + \left( {{a^2}{\omega ^2} + {b^2}{\omega ^4} + 2ab{w^3}} \right) + \left( {{a^2}{\omega ^4} + {b^2}{\omega ^2} + 2ab{\omega ^3}} \right)\]
Opening the parenthesis, so that same type of values can be added, and we get:
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {a^2} + {b^2} + 2ab + {a^2}{\omega ^2} + {b^2}{\omega ^4} + 2ab{w^3} + {a^2}{\omega ^4} + {b^2}{\omega ^2} + 2ab{\omega ^3}\]
Separating out the value’s having \[{a^2}\] on one side and then applying a parenthesis and similarly for \[{b^2}\] and \[2ab\]:
\[ \Rightarrow {x^2} + {y^2} + {z^2} = \left( {{a^2} + {a^2}{\omega ^4} + {a^2}{\omega ^2}} \right) + \left( {{b^2} + {b^2}{\omega ^2} + {b^2}{\omega ^4}} \right) + \left( {2ab + 2ab{w^3} + 2ab{\omega ^3}} \right)\]
Taking \[{a^2}\], \[{b^2}\] and \[2ab\] common from their respective brackets:
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {a^2}\left( {1 + {\omega ^4} + {\omega ^2}} \right) + {b^2}\left( {1 + {\omega ^2} + {\omega ^4}} \right) + 2ab\left( {1 + {w^3} + {\omega ^3}} \right)\]
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {a^2}\left( {1 + {\omega ^4} + {\omega ^2}} \right) + {b^2}\left( {1 + {\omega ^2} + {\omega ^4}} \right) + 2ab\left( {1 + 2{w^3}} \right)\]
From the law of radicals, \[{\omega ^4}\] can be splitted and written as \[{\omega ^4} = {\omega ^3}.\omega \], so writing it in the above equation, and we get:
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {a^2}\left( {1 + {\omega ^3}.\omega + {\omega ^2}} \right) + {b^2}\left( {1 + {\omega ^2} + {\omega ^3}.\omega } \right) + 2ab\left( {1 + 2{w^3}} \right)\] ….(2)
Since, it’s already given that omega ($\omega $) is an imaginary cube root of unity that implies \[{\omega ^3} = 1\].
So, taking \[{\omega ^3} = 1\] in equation 2, we get:
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {a^2}\left( {1 + \omega + {\omega ^2}} \right) + {b^2}\left( {1 + \omega + {\omega ^2}} \right) + 2ab\left( {1 + 2} \right)\]
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {a^2}\left( {1 + \omega + {\omega ^2}} \right) + {b^2}\left( {1 + \omega + {\omega ^2}} \right) + 2ab\left( 3 \right)\]
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {a^2}\left( {1 + \omega + {\omega ^2}} \right) + {b^2}\left( {1 + \omega + {\omega ^2}} \right) + 6ab\] ……(3)
Since, we know that the sum of the three cube roots of infinity is zero, that is:
\[\left( {1 + \omega + {\omega ^2}} \right) = 0\]
Substituting this value in equation 3, we get:
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {a^2}\left( 0 \right) + {b^2}\left( 0 \right) + 6ab\]
\[ \Rightarrow {x^2} + {y^2} + {z^2} = 6ab\]
Therefore, the value of \[{x^2} + {y^2} + {z^2} = 6ab\].
Hence, Option 1 is correct.

Note: According to the law of radicals, the powers can be added if they have same bases and they are getting multiplied, for example ${p^2}.{p^5} = {p^{2 + 5}} = {p^7}$.And, similarly for division the powers are getting subtracted. It’s important to remember the important rules of omega or the steps to prove the theorems of omega.