Question

If \[x + \dfrac{1}{x} = \sqrt 3 \] then find the value of \[{x^{17}} + \dfrac{1}{{{x^{17}}}}\] ?

Answer 1

Hint: In order to solve this question, we will try to make \[{x^{17}} + \dfrac{1}{{{x^{17}}}}\] .For this, first of all we will take the square and cube of the given equation and multiply both the equations. And hence we will get the value of \[{x^5} + \dfrac{1}{{{x^5}}}\] . After that we will find out the value of \[{x^7} + \dfrac{1}{{{x^7}}}\] by multiplying \[{x^5} + \dfrac{1}{{{x^5}}}\] and \[{x^2} + \dfrac{1}{{{x^2}}}\] . Then we will find out the value of \[{x^{10}} + \dfrac{1}{{{x^{10}}}}\] by taking square of \[{x^5} + \dfrac{1}{{{x^5}}}\] . And finally, we will find the value of \[{x^{17}} + \dfrac{1}{{{x^{17}}}}\] by multiplying the values of \[{x^7} + \dfrac{1}{{{x^7}}}\] and \[{x^{10}} + \dfrac{1}{{{x^{10}}}}\] .Hence, we will get the required result.

Complete step by step answer:
We have given, \[x + \dfrac{1}{x} = \sqrt 3 {\text{ }} - - - \left( 1 \right)\]
Taking square on both sides, we get
\[{\left( {x + \dfrac{1}{x}} \right)^2} = {\left( {\sqrt 3 } \right)^2}\]
\[ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 3\]
\[ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} = 1{\text{ }} - - - \left( 2 \right)\]
Now taking cube on both sides of equation \[\left( 1 \right)\] we get
\[{\left( {x + \dfrac{1}{x}} \right)^3} = {\left( {\sqrt 3 } \right)^3}\]
\[ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3 \cdot x \cdot \dfrac{1}{x}\left( {x + \dfrac{1}{x}} \right) = 3\sqrt 3 \]

On substituting value from equation \[\left( 1 \right)\] we get
\[ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \]
On solving, we get
\[ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 0{\text{ }} - - - \left( 3 \right)\]
Now on multiplying equation \[\left( 2 \right)\] and \[\left( 3 \right)\] we get
\[\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)\left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) = 0\]
\[ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} + x + \dfrac{1}{x} = 0\]

On substituting value from equation \[\left( 1 \right)\] we get
\[ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} = - \sqrt 3 {\text{ }} - - - \left( 4 \right)\]
Now multiplying equation \[\left( 2 \right)\] and \[\left( 4 \right)\] we get
\[ \Rightarrow \left( {{x^5} + \dfrac{1}{{{x^5}}}} \right)\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = \left( { - \sqrt 3 } \right)\left( 1 \right)\]
\[ \Rightarrow {x^7} + \dfrac{1}{{{x^7}}} + {x^3} + \dfrac{1}{{{x^3}}} = - \sqrt 3 \]
On substituting value from equation \[\left( 3 \right)\] we get
\[ \Rightarrow {x^7} + \dfrac{1}{{{x^7}}} = - \sqrt 3 {\text{ }} - - - \left( 5 \right)\]

Now taking square on both the sides of the equation \[\left( 4 \right)\] we get
\[{\left( {{x^5} + \dfrac{1}{{{x^5}}}} \right)^2} = {\left( { - \sqrt 3 } \right)^2}\]
\[ \Rightarrow {x^{10}} + \dfrac{1}{{{x^{10}}}} + 2 = 3\]
\[ \Rightarrow {x^{10}} + \dfrac{1}{{{x^{10}}}} = 1{\text{ }} - - - \left( 6 \right)\]
Now multiplying equation \[\left( 5 \right)\] and \[\left( 6 \right)\] we get
\[ \Rightarrow \left( {{x^7} + \dfrac{1}{{{x^7}}}} \right)\left( {{x^{10}} + \dfrac{1}{{{x^{10}}}}} \right) = \left( { - \sqrt 3 } \right)\left( 1 \right)\]
\[ \Rightarrow {x^{17}} + \dfrac{1}{{{x^{17}}}} + {x^3} + \dfrac{1}{{{x^3}}} = - \sqrt 3 \]
On substituting value from equation \[\left( 3 \right)\] we get
\[ \therefore {x^{17}} + \dfrac{1}{{{x^{17}}}} = - \sqrt 3 \]

Hence, If \[x + \dfrac{1}{x} = \sqrt 3 \] then the value of \[{x^{17}} + \dfrac{1}{{{x^{17}}}} = - \sqrt 3 \].

Note: Alternative way to solve this problem:
We have given:
\[x + \dfrac{1}{x} = \sqrt 3 {\text{ }} - - - \left( 1 \right)\]
Squaring both sides, we get
\[ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 3\]
Subtracting \[3\] from both sides, we get
\[ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} - 1 = 0\]
Now multiply both sides by \[{x^4} + {x^2}\] we get
\[\left( {{x^4} + {x^2}} \right)\left( {{x^2} + \dfrac{1}{{{x^2}}} - 1} \right) = 0\]
\[ \Rightarrow {x^6} + {x^4} + {x^2} + 1 - {x^4} - {x^2} = 0\]
On cancelling the like terms, we get
\[ \Rightarrow {x^6} + 1 = 0\]
This means, \[x\] is the sixth root of \[ - 1\]
So, \[{x^{17}} + \dfrac{1}{{{x^{17}}}}\] can be written as
\[{x^{17}} + \dfrac{1}{{{x^{17}}}} = \dfrac{{{x^{18}}}}{x} + \dfrac{x}{{{x^{18}}}}\]
As \[{x^6} = - 1{\text{ }} \Rightarrow {x^{18}} = {\left( { - 1} \right)^3}\]
Therefore, we get
\[ \Rightarrow {x^{17}} + \dfrac{1}{{{x^{17}}}} = \dfrac{{{{\left( { - 1} \right)}^3}}}{x} + \dfrac{x}{{{{\left( { - 1} \right)}^3}}}\]
\[ \Rightarrow {x^{17}} + \dfrac{1}{{{x^{17}}}} = \dfrac{{ - 1}}{x} - \dfrac{x}{1}\]
Taking \[ - 1\] common, we get
\[ \Rightarrow {x^{17}} + \dfrac{1}{{{x^{17}}}} = - \left( {x + \dfrac{1}{x}} \right)\]
using equation \[\left( 1 \right)\] we get
\[ \Rightarrow {x^{17}} + \dfrac{1}{{{x^{17}}}} = - \sqrt 3 \]
Hence, we get the required result.