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Question: If \({x^2} + x + 1 = 0\) and \(\omega \ne 1\) then the value of \({\left( {1 + 3\omega + 3{\omega ^2...

If x2+x+1=0{x^2} + x + 1 = 0 and ω1\omega \ne 1 then the value of (1+3ω+3ω2)8+(3+ω+3ω2)8+(3+3ω+ω2)8{\left( {1 + 3\omega + 3{\omega ^2}} \right)^8} + {\left( {3 + \omega + 3{\omega ^2}} \right)^8} + {\left( {3 + 3\omega + {\omega ^2}} \right)^8} equals

Explanation

Solution

Given an expression. We have to compute the value of the expression. First, we will apply the properties for the cube root of unity to the expression. Then, substitute the value of cube root unity into the expression. Then, we will simplify the expression by taking out the common terms. Then, substitute the value of the reduced expression.

Formula used: The properties of cube root of the unity is defined as:
ω3=1{\omega ^3} = 1 and 1+ω+ω2=01 + \omega + {\omega ^2} = 0

Complete step-by-step solution:
Let ω\omega be the cube root of the unity. Now, simplify the expression by taking out the common terms in each of the components of the expression.
(1+3(ω+ω2))8+(ω+3(1+ω2))8+(3(1+ω)+ω2)8\Rightarrow {\left( {1 + 3\left( {\omega + {\omega ^2}} \right)} \right)^8} + {\left( {\omega + 3\left( {1 + {\omega ^2}} \right)} \right)^8} + {\left( {3\left( {1 + \omega } \right) + {\omega ^2}} \right)^8}
Now, use the property 1+ω+ω2=01 + \omega + {\omega ^2} = 0 to compute the values of the expression.
ω+ω2=1\Rightarrow \omega + {\omega ^2} = - 1
1+ω2=ω\Rightarrow 1 + {\omega ^2} = - \omega
1+ω=ω2\Rightarrow 1 + \omega = - {\omega ^2}
Now, we will substitute the values into the expression.
(1+3(1))8+(ω+3(ω))8+(3(ω2)+ω2)8\Rightarrow {\left( {1 + 3\left( { - 1} \right)} \right)^8} + {\left( {\omega + 3\left( { - \omega } \right)} \right)^8} + {\left( {3\left( { - {\omega ^2}} \right) + {\omega ^2}} \right)^8}
On simplifying the expression, we get:
(13)8+(ω3ω)8+(3ω2+ω2)8\Rightarrow {\left( {1 - 3} \right)^8} + {\left( {\omega - 3\omega } \right)^8} + {\left( { - 3{\omega ^2} + {\omega ^2}} \right)^8}
Now, we will combine like terms to simplify the expression.
(2)8+(2ω)8+(2ω2)8\Rightarrow {\left( { - 2} \right)^8} + {\left( { - 2\omega } \right)^8} + {\left( { - 2{\omega ^2}} \right)^8}
Since the exponent of the terms is even, thus the expression becomes positive. Now, we will remove the negative signs before each term of the expression.
(2)8+(2ω)8+(2ω2)8\Rightarrow {\left( 2 \right)^8} + {\left( {2\omega } \right)^8} + {\left( {2{\omega ^2}} \right)^8}
Now, take out the (2)8{\left( 2 \right)^8} as a common term.
28(1+ω8+(ω2)8)\Rightarrow {2^8}\left( {1 + {\omega ^8} + {{\left( {{\omega ^2}} \right)}^8}} \right)
Now, we will substitute the value 1+ω+ω2=01 + \omega + {\omega ^2} = 0 to the expression.
28(0)\Rightarrow {2^8}\left( 0 \right)
Now, multiply the terms to further reduce the expression, we get:
0\Rightarrow 0
Hence, the value of (1+3ω+3ω2)8+(3+ω+3ω2)8+(3+3ω+ω2)8{\left( {1 + 3\omega + 3{\omega ^2}} \right)^8} + {\left( {3 + \omega + 3{\omega ^2}} \right)^8} + {\left( {3 + 3\omega + {\omega ^2}} \right)^8} is 00

Thus, the correct answer is option B.

Note: In such types of questions the students mainly don't get an approach on how to solve it. In such types of questions students mainly forget to apply the properties of the cube root of unity. Students may get confused while finding the values using the relation 1+ω+ω2=01 + \omega + {\omega ^2} = 0.