If $x^2-5 x+1=0$, then the value of $\frac{x^6+x^4+x^2+1}{5 x^3}=$ ?
Answer & explanation
Correct answer: option 3
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