Question

if x^2-x-1 =0 then what is the value of x^8 + 1/x^8

Answer 1

$\textbf{Answer}$


x^8 + 1/x^8 = ?

$\textbf{We are given,}$
x^2 - x - 1 = 0 -------(1)
$\textbf{Dividing}$ both $\textbf{sides of equation(1) by x,}$
x^2/x - x/x - 1/x = 0/x
x^2 - 1/x - 1 = 0

=> $\textbf{(x - 1/x) = 1}$
$\textbf{Squaring both sides,}$
(x - 1/x)^2 = (1)^2
=> x^2 + 1/x^2 - 2(x)(1/x) = 1

=> (x^2 + 1/x^2) = 3
$\textbf{Squaring both sides,}$
(x^2 + 1/x^2)^2 = (3)^2
=> x^4 + 1/x^4 + 2(x^4)(1/x^4) = 9
=> x^4 + 1/x^4 + 2 = 9

=> (x^4 + 1/x^4) = 7
$\textbf{Squaring both sides,}$
x^8 + 1/x^8 + 2(x^8)(1/x^8) = (7)^2
=> x^8 + 1/x^8 + 2 = 49
=> x^8 + 1/x^8 = 49 - 2

=> x^8 + 1/x^8 = 47 $\textbf{(Solution)}$


$\textbf{Hope My Answer Helped}$
$\textbf{Thanks}$

Answer 2

Given:

• x² - x - 1 = 0

TO FIND:

• x⁸ + (1/x⁸)

SOLUTION:

x² - x - 1 = 0

x² - 1 = 1x

x²/x - 1/x = 1

x - 1/x = 1

S.O.B.S

(x - 1/x)² = 1²

x² - 2(x)(1/x) + 1/x² = 1

x² + 1/x² = 3

SOBS

(x² + 1/x²)² = 3²

x⁴ + 2(x²)(1/x²) + 1/x⁴ = 9

x⁴ + 1/x⁴ = 7

SOBS

(x⁴ + 1/x⁴)² = 7²

x⁸ + 2(x⁴)(1/x⁴) + 1/x⁸ = 49

x⁸ + 1/x⁸ = 47. ( Answer )

Hence, x⁸ + 1/x⁸ = 47

USED FORMULAS:

• ( a + b )² = a² + 2ab + b²
• ( a - b )² = a² - 2ab + b²