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Question: If \({x^2} + x + 1 = 0\), then the value of \({\left( {x + \dfrac{1}{x}} \right)^2} + {\left( {{x^2}...

If x2+x+1=0{x^2} + x + 1 = 0, then the value of (x+1x)2+(x2+1x2)2+...+(x27+1x27)2{\left( {x + \dfrac{1}{x}} \right)^2} + {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} + ... + {\left( {{x^{27}} + \dfrac{1}{{{x^{27}}}}} \right)^2} is
A. 27
B. 72
C. 45
D. 54

Explanation

Solution

Recall that x3−1=(x−1)(x2+x+1){x^3} - 1 = (x - 1)({x^2} + x + 1) , therefore, x is a cube−root of unity in the given equation. If ω\omega is a cube−root of unity, then w3=1⇒w2=1w{w^3} = 1 \Rightarrow {w^2} = \dfrac{1}{w} and w2+w+1=0⇒w2+w=−1{w^2} + w + 1 = 0 \Rightarrow {w^2} + w = - 1 . Reduce the higher powers to ω2{\omega ^2} and ω\omega , and evaluate. Observe that there are 27 groups of squared terms and their values will repeat in a fixed cycle.

Complete step by step solution:
From x2+x+1=0{x^2} + x + 1 = 0 , we know that the roots are ω\omega and ω2{\omega ^2} such that w2+w+1=0{w^2} + w + 1 = 0 and ω3=1{\omega ^3} = 1 .
It also means that w2=1w{w^2} = \dfrac{1}{w} . Therefore, w4=w3×w=w{w^4} = {w^3} \times w = w etc.
Now, the expression (x+1x)2+(x2+1x2)2+...\+(x27+1x27)2{\left( {x + \dfrac{1}{x}} \right)^2} + {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} + ... \+ {\left( {{x^{27}} + \dfrac{1}{{{x^{27}}}}} \right)^2} can be written as:
= (ω+ω2)2+(ω2+ω)2+(ω3+ω3)2+(ω+ω2)2{\left( {\omega + {\omega ^2}} \right)^2} + {\left( {{\omega ^2} + \omega } \right)^2} + {\left( {{\omega ^3} + {\omega ^3}} \right)^2} + {\left( {\omega + {\omega ^2}} \right)^2} + ... + (ω3+ω3)2{\left( {{\omega ^3} + {\omega ^3}} \right)^2} (27 terms)
Substituting ω2+ω=−1{\omega ^2} + \omega = - 1 and ω3=1{\omega ^3} = 1, we get:
= (−1)2+(−1)2+(4)2+(−1)2{( - 1)^2} + {( - 1)^2} + {(4)^2} + {( - 1)^2} + ... + (4)2{(4)^2} (27 terms)
Simplifying squares, we get,
= 1 + 1 + 4 + 1 + ... + 4 (27 terms)
Making triplets and combine, we get,
= (1 + 1 + 4) × 9
= 54
The correct answer is option D. 54.

Note: The question can also be solved as follows:
x2+x+1=0{x^2} + x + 1 = 0
Dividing by x, we get:
⇒x+1+1x=0\Rightarrow x + 1 + \dfrac{1}{x} = 0
⇒x+1x=−1\Rightarrow x + \dfrac{1}{x} = - 1 ... (1)
On squaring both sides, we get,
And, (x+1x)2=(−1)2{\left( {x + \dfrac{1}{x}} \right)^2} = {( - 1)^2}
Expanding Identity, we get,
⇒x2+1x2+2=1\Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 1
⇒x2+1x2=−1\Rightarrow {x^2} + \dfrac{1}{{{x^2}}} = - 1 ... (2)
And, (x+1x)3=(−1)3{\left( {x + \dfrac{1}{x}} \right)^3} = {( - 1)^3}
Expanding identity, we get,
⇒x3+1x3+3(x×1x)(x+1x)=−1\Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {x \times \dfrac{1}{x}} \right)\left( {x + \dfrac{1}{x}} \right) = - 1
⇒x3+1x3+3(−1)=−1\Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3( - 1) = - 1
⇒x3+1x3=2\Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 2 ... (3)
And,
x4+1x4=(x2+1x2)2−2=1−2=−1{x^4} + \dfrac{1}{{{x^4}}} = {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} - 2 = 1 - 2 = - 1 ...(4)
And,
(x2+1x2)(x3+1x3)=x5+1x+x+1x5\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)\left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) = {x^5} + \dfrac{1}{x} + x + \dfrac{1}{{{x^5}}}
⇒ x5+1x5=(−1)×2+1=−1{x^5} + \dfrac{1}{{{x^5}}} = ( - 1) \times 2 + 1 = - 1 ... (5)
And so on.
But this method is not advised because it is hard to generalize a pattern here.
The three cube−roots of the unity (1) are: w=−1+i32w = \dfrac{{ - 1 + i\sqrt 3 }}{2}, ω2=−1−i32{\omega ^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2} and 1.