Question

refer to the attachhment​

Answer 1

Answer = 123

Step by step Explanation

Given

$\sf x²-3x+1=0$

To Find

$\sf x⁵+ \dfrac{1}{ {x}^{5} }$

$\sf \: \: \: \: \: → \: \: x²-3x+1=0$

On Dividing Both side's by x

$\sf \: \: \: \: \: → \: \: \: \dfrac{ {x}^{2} }{x} - \dfrac{3x}{x} + \dfrac{1}{x} = \dfrac{0}{3}$

$\sf \: \: \: \: \: → \: \: \: x - 3 + \dfrac{1}{x} =0$

Adding both sides by 3

$\sf \: \: \: \: \: → \: \: \: x \cancel{ - 3 }+ \dfrac{1}{x} \cancel{ + 3}=0 + 3$

$\sf \: \: \: \: \: → \: \: \: x + \dfrac{1}{x} =3 \: \: \: \: \: - - - - (1)$

Now

On Squaring Both Side's

${ \sf \: \: \: \: \: → \: \: \: \left(x + \dfrac{1}{x} \right)^{2} = {3}^{2} }$

${ \sf \: \: \: \: \: → \: \: \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times \cancel{x }\times \dfrac{1}{ \cancel{x}} = 9 }$

Subtracting both sides by 2

${ \sf \: \: \: \: \: → \: \: \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \ - 2 = 9 - 2}$

${ \sf \: \: \: \: \: → \: \: \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 7 \: \: \: \: \: - - - - (2)}$

Now, Again

On Cubing both sides in Equation 1

${ \sf \: \: \: \: \: → \: \: \: \left( x + \dfrac{1}{x} \right)^{3} = {3}^{3} }$

${ \sf \: \: \: \: \: → \: \: \: \ {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x}(x + \dfrac{1}{x} )= {3}^{3} }$

${ \sf \: \: \: \: \: → \: \: \: \ {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times3= 27 }$

${ \sf \: \: \: \: \: → \: \: \: \ {x}^{3} + \dfrac{1}{ {x}^{3} } + 9= 27 }$

Subtracting both sides by 9

${ \sf \: \: \: \: \: → \: \: \: \ {x}^{3} + \dfrac{1}{ {x}^{3} } + 9 - 9= 27 - 9 }$

${ \sf \: \: \: \: \: → \: \: \: \ {x}^{3} + \dfrac{1}{ {x}^{3} } = 18 \: \: \: \: \: - - - - (3) }$

Now

Multiplying Equation 1 and 2 , we get

${ \sf \: \: \: \: \: → \: \: \: \left( {x}^{2} + \dfrac{1}{ {x}^{2} } \right) \left({x}^{3} + \dfrac{1}{ {x}^{3} } \right) = 7 \times 18 }$

${ \sf \: \: \: \: \: → \: \: \: {x}^{2} \left({x}^{3} + \dfrac{1}{ {x}^{3} } \right) + \dfrac{1}{ {x}^{2} }\left({x}^{3} + \dfrac{1}{ {x}^{3} } \right) = 126 }$

${ \sf \: \: \: \: \: → \: \: \: {x}^{5} + \dfrac{1}{ {x}^{5} } + x + \dfrac{1}{ x } = 126 }$

Putting the value of x+1/x

${ \sf \: \: \: \: \: → \: \: \: {x}^{5} + \dfrac{1}{ {x}^{5} } + 3 = 126 }$

Subtracting both sides by 3

${ \sf \: \: \: \: \: → \: \: \: {x}^{5} + \dfrac{1}{ {x}^{5} } + 3 - 3 = 126 - 3 }$

${ \sf \: \: \: \: \: → \: \: \: {x}^{5} + \dfrac{1}{ {x}^{5} } + = 123}$

Hance (b) is the correct option.