Question

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

Answer 1

Hint:Recall that ${x^3} - 1 = (x - 1)({x^2} + x + 1)$ , therefore, x is a cube−root of unity in the given equation. If $\omega $ is a cube−root of unity, then ${w^3} = 1 \Rightarrow {w^2} = \dfrac{1}{w}$ and ${w^2} + w + 1 = 0 \Rightarrow {w^2} + w = - 1$ . Reduce the higher powers to ${\omega ^2}$ and $\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 ${x^2} + x + 1 = 0$ , we know that the roots are $\omega $ and ${\omega ^2}$ such that ${w^2} + w + 1 = 0$ and ${\omega ^3} = 1$ .
It also means that ${w^2} = \dfrac{1}{w}$ . Therefore, ${w^4} = {w^3} \times w = w$ etc.
Now, the expression ${\left( {x + \dfrac{1}{x}} \right)^2} + {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} + ...
+ {\left( {{x^{27}} + \dfrac{1}{{{x^{27}}}}} \right)^2}$ can be written as:
= ${\left( {\omega + {\omega ^2}} \right)^2} + {\left( {{\omega ^2} + \omega } \right)^2} + {\left(
{{\omega ^3} + {\omega ^3}} \right)^2} + {\left( {\omega + {\omega ^2}} \right)^2}$ + ... + ${\left(
{{\omega ^3} + {\omega ^3}} \right)^2}$ (27 terms)
Substituting ${\omega ^2} + \omega = - 1$ and ${\omega ^3} = 1$, we get:
= ${( - 1)^2} + {( - 1)^2} + {(4)^2} + {( - 1)^2}$ + ... + ${(4)^2}$ (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:
${x^2} + x + 1 = 0$
Dividing by x, we get:
$ \Rightarrow x + 1 + \dfrac{1}{x} = 0$
$ \Rightarrow x + \dfrac{1}{x} = - 1$ ... (1)
On squaring both sides, we get,
And, ${\left( {x + \dfrac{1}{x}} \right)^2} = {( - 1)^2}$
Expanding Identity, we get,
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 1$
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} = - 1$ ... (2)
And, ${\left( {x + \dfrac{1}{x}} \right)^3} = {( - 1)^3}$
Expanding identity, we get,
$ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {x \times \dfrac{1}{x}} \right)\left( {x + \dfrac{1}{x}}
\right) = - 1$
$ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3( - 1) = - 1$
$ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 2$ ... (3)
And,
${x^4} + \dfrac{1}{{{x^4}}} = {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} - 2 = 1 - 2 = - 1$ ...(4)
And,
$\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)\left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) = {x^5} + \dfrac{1}{x} + x + \dfrac{1}{{{x^5}}}$
⇒ ${x^5} + \dfrac{1}{{{x^5}}} = ( - 1) \times 2 + 1 = - 1$ ... (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: $w = \dfrac{{ - 1 + i\sqrt 3 }}{2}$, ${\omega ^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}$ and 1.