If $x + \frac{1}{x} = 5, x ≠ 0 $ then the value of $\frac{x^4+\frac{1}{x^2}}{x^2-3x+1}$ is equal to :

55

$x + \frac{1}{x} = 5, x ≠ 0 $

then the value of $\frac{x^4+\frac{1}{x^2}}{x^2-3x+1}$

We know that,

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

Divide numerator and denominator by x we get,

We can write $\frac{x^3+\frac{1}{x^3}}{x-3+\frac{1}{x}}$

= $x^3 +\frac{1}{x^3}$ = 5³ - 3 × 5 = 110

Put this value in $\frac{x^3+\frac{1}{x^3}}{x-3+\frac{1}{x}}$

= $\frac{110}{5-3}$ = 55