If $x^2-5 x+1=0$, then the value of $\frac{x^6+x^4+x^2+1}{5 x^3}=$ ?

23

We have,

$x^2-5 x+1=0$

We can write the above equation as,

x + \(\frac{1}{x}\) = 5

So, x³ + \(\frac{1}{x^3}\) = 5³ - 5 × 3 = 110

We have to find the value of $\frac{x^6+x^4+x^2+1}{5 x^3}$

Taking x³ as common from both the numerator and denominator we get,

= (x³ + \(\frac{1}{x^3}\) + x + \(\frac{1}{x}\)) / 5

= (110 + 5) / 5

= 23