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Question: If omega (\(\omega \)) is an imaginary cube root of unity and \(x = a + b\), \(y = a\omega+ b{\omega...

If omega (ω\omega ) is an imaginary cube root of unity and x=a+bx = a + b, y=aω+bω2y = a\omega+ b{\omega ^2}, z=aω2+bωz = a{\omega ^2} + b\omega , then x2+y2+z2{x^2} + {y^2} + {z^2} is equal to
1). 6ab6ab
2). 3ab3ab
3). 6a2b26{a^2}{b^2}
4). 3a2b23{a^2}{b^2}

Explanation

Solution

In order to find the value of x2+y2+z2{x^2} + {y^2} + {z^2}, substitute the value of the variables given that is x2,y2,z2{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. (x+y)2=x2+y2+2xy{\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy
2. (1+ω+ω2)=0\left( {1 + \omega + {\omega ^2}} \right) = 0
3. ω3=1{\omega ^3} = 1

Complete step-by-step solution:
We are given with three equations, that are stated as:
x=a+bx = a + b
y=aω+bω2y = a\omega + b{\omega ^2}
z=aω2+bωz = a{\omega ^2} + b\omega
We need to find the value of x2+y2+z2{x^2} + {y^2} + {z^2}.
So, substituting the value of xx, yy and zz in the equation x2+y2+z2{x^2} + {y^2} + {z^2}, and we get:
x2+y2+z2=(a+b)2+(aω+bω2)2+(aω2+bω)2{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
(x+y)2=x2+y2+2xy{\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:
⇒x2+y2+z2=(a2+b2+2ab)+(a2ω2+b2ω4+2abw3)+(a2ω4+b2ω2+2abω3)\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:
⇒x2+y2+z2=a2+b2+2ab+a2ω2+b2ω4+2abw3+a2ω4+b2ω2+2abω3\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 a2{a^2} on one side and then applying a parenthesis and similarly for b2{b^2} and 2ab2ab:
⇒x2+y2+z2=(a2+a2ω4+a2ω2)+(b2+b2ω2+b2ω4)+(2ab+2abw3+2abω3)\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 a2{a^2}, b2{b^2} and 2ab2ab common from their respective brackets:
⇒x2+y2+z2=a2(1+ω4+ω2)+b2(1+ω2+ω4)+2ab(1+w3+ω3)\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)
⇒x2+y2+z2=a2(1+ω4+ω2)+b2(1+ω2+ω4)+2ab(1+2w3)\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, ω4{\omega ^4} can be splitted and written as ω4=ω3.ω{\omega ^4} = {\omega ^3}.\omega , so writing it in the above equation, and we get:
⇒x2+y2+z2=a2(1+ω3.ω+ω2)+b2(1+ω2+ω3.ω)+2ab(1+2w3)\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 (ω\omega ) is an imaginary cube root of unity that implies ω3=1{\omega ^3} = 1.
So, taking ω3=1{\omega ^3} = 1 in equation 2, we get:
⇒x2+y2+z2=a2(1+ω+ω2)+b2(1+ω+ω2)+2ab(1+2)\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)
⇒x2+y2+z2=a2(1+ω+ω2)+b2(1+ω+ω2)+2ab(3)\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)
⇒x2+y2+z2=a2(1+ω+ω2)+b2(1+ω+ω2)+6ab\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:
(1+ω+ω2)=0\left( {1 + \omega + {\omega ^2}} \right) = 0
Substituting this value in equation 3, we get:
⇒x2+y2+z2=a2(0)+b2(0)+6ab\Rightarrow {x^2} + {y^2} + {z^2} = {a^2}\left( 0 \right) + {b^2}\left( 0 \right) + 6ab
⇒x2+y2+z2=6ab\Rightarrow {x^2} + {y^2} + {z^2} = 6ab
Therefore, the value of x2+y2+z2=6ab{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 p2.p5=p2+5=p7{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.