Solveeit Logo
Login

Question

Question: If cube root of unity is 1, \(\omega ,{\omega ^2}\)then what will be the value of \({\omega ^2}{\lef...

If cube root of unity is 1, ω,ω2\omega ,{\omega ^2}then what will be the value of ω2(1+ω)3−(1+ω2)ω{\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega .
(a) 1 (b) - 1 (c) i (d) 0  (a){\text{ 1}} \\\ (b){\text{ - 1}} \\\ (c){\text{ i}} \\\ (d){\text{ 0}} \\\

Explanation

Solution

Hint: Use the property of the cube root of unity that 1+ω+ω2=01 + \omega + {\omega ^2} = 0and ω3=1{\omega ^3} = 1 to simplify the given expression. The simplification needs to be done by keeping one thing in mind that we have to make terms form the above two equations as its value is known.

Complete step-by-step answer:
The given expression is
ω2(1+ω)3−(1+ω2)ω{\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega …………………. (1)
Now it is given that 1,ω,ω21,\omega ,{\omega ^2} are the cube of unity.
Then according to the property of cube root of unity we have,
⇒1+ω+ω2=0...............(2),ω3=1............(3)\Rightarrow 1 + \omega + {\omega ^2} = 0...............\left( 2 \right),{\omega ^3} = 1............\left( 3 \right)
Now first simplify the equation (1) according to property(a+b)3=a3+b3+3ab2+3a2b{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3a{b^2} + 3{a^2}b, we have,
⇒ω2(1+ω)3−(1+ω2)ω=ω2(1+ω3+3ω2+3ω)−ω−ω3\Rightarrow {\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega = {\omega ^2}\left( {1 + {\omega ^3} + 3{\omega ^2} + 3\omega } \right) - \omega - {\omega ^3}
Now again simplify we have,
⇒ω2(1+ω)3−(1+ω2)ω=ω2+ω2ω3+3ω2ω2+3ω3−ω−ω3\Rightarrow {\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega = {\omega ^2} + {\omega ^2}{\omega ^3} + 3{\omega ^2}{\omega ^2} + 3{\omega ^3} - \omega - {\omega ^3}
Now simplify the above equation using equation (3) we have,
⇒ω2(1+ω)3−(1+ω2)ω=ω2+ω2(1)+3ω(1)+3(1)−ω−(1)\Rightarrow {\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega = {\omega ^2} + {\omega ^2}\left( 1 \right) + 3\omega \left( 1 \right) + 3\left( 1 \right) - \omega - \left( 1 \right)
Now simplify the above equation we have,
⇒ω2(1+ω)3−(1+ω2)ω=2ω2+2ω+2\Rightarrow {\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega = 2{\omega ^2} + 2\omega + 2
⇒ω2(1+ω)3−(1+ω2)ω=2(ω2+ω+1)\Rightarrow {\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega = 2\left( {{\omega ^2} + \omega + 1} \right)
Now from equation (2) we have,
⇒ω2(1+ω)3−(1+ω2)ω=2(0)=0\Rightarrow {\omega ^2}{\left( {1 + \omega } \right)^3} - \left( {1 + {\omega ^2}} \right)\omega = 2\left( 0 \right) = 0
So the required solution of the given equation is zero.
Hence option (d) is correct.

Note: Cube root of unity refers to the cube roots of 1 that is (1)13=(1,ω,ω2){\left( 1 \right)^{\dfrac{1}{3}}} = \left( {1,\omega, {\omega ^2}} \right). It is always advisable to have a good grasp over the basic identities involving cube root of unity as it helps saving a lot of time, and helps in simplification of the problem statement.