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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 $54$

Last updated date: 17th Jul 2024
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348.9k+ views
Hint: To get the value of the given equation we will use the root of unity concept. Firstly we will let the variable in question be equal to the root of unity. Then by using the value and relation of the roots of unity we will put the values in the equation. Finally we will simplify the equation to get the desired answer.

Complete step by step answer:
The equation is given as below:
${{x}^{2}}+x+1=0$…….$\left( 1 \right)$
So it will have roots as:
$x=\omega ,{{\omega }^{2}}$
Where $\omega ,{{\omega }^{2}}$ are cube roots of unity.
On substituting above value in equation (1) we get,
\[{{\omega }^{2}}+\omega +1=0\]……$\left( 2 \right)$
Also we know cube of an imaginary cube root is unity so,
${{\omega }^{3}}=1$…..$\left( 3 \right)$
So fro above value we can get:
  & {{\omega }^{2}}.\omega =1 \\
 & {{\omega }^{2}}=\dfrac{1}{\omega } \\
\end{align}$…..$\left( 4 \right)$
${{\omega }^{3n}}=1$…….$\left( 5 \right)$
We have to find the value of below equation:
 ${{\left( x+\dfrac{1}{x} \right)}^{2}}+{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}+.....+{{\left( {{x}^{27}}+\dfrac{1}{{{x}^{27}}} \right)}^{2}}$
Using equation (2) - (5) we will solve above equation as:
  & \Rightarrow {{\left( \omega +\dfrac{{{\omega }^{3}}}{\omega } \right)}^{2}}+{{\left( {{\omega }^{2}}+\dfrac{{{\omega }^{3}}}{{{\omega }^{2}}} \right)}^{2}}+{{\left( {{\omega }^{3}}+\dfrac{{{\omega }^{3}}}{{{\omega }^{3}}} \right)}^{2}}.....+{{\left( {{\omega }^{27}}+\dfrac{{{\omega }^{3}}}{{{\omega }^{27}}} \right)}^{2}} \\
 & \Rightarrow {{\left( \omega +{{\omega }^{2}} \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( {{\omega }^{3}}+1 \right)}^{2}}.....+{{\left( {{\omega }^{27}}+\dfrac{1}{{{\omega }^{24}}} \right)}^{2}} \\
 & \Rightarrow {{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 1+1 \right)}^{2}}....+{{\left( {{\omega }^{27}} +\dfrac{1}{1} \right)}^{2}} \\
 & \Rightarrow {{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}....+{{\left( 1+1 \right)}^{2}} \\
 & \Rightarrow {{\left( -1 \right)}^{2}}+ {{\left( -1 \right)}^{2}}+......{{\left( 2 \right)}^{2}} \\
This simplified further give,
${{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{2}^{2}}+.....{{\left( 2 \right)}^{2}}$
So from the above value we get that we are getting two terms same then a different term it will go on like this
  & \Rightarrow 18{{\left( -1 \right)}^{2}}+9{{\left( 2 \right)}^{2}} \\
 & \Rightarrow 18+9\times 4 \\
 & \Rightarrow 18+36 \\
 & \Rightarrow 54 \\
So value is obtained as 54.

 So, the correct answer is “Option D”.

Note: A number which when raised to the power 3 gives the answer as 1 is known as cube root of unity. It is widely used in many branches of mathematics. There are three roots of unity in which two are complex roots and one is a real root. Some properties of the cube root of unity are that when one imaginary root is squared it gives another root of unity. When two complex roots are multiplied the answer comes as 1.