If $ x + \frac{1}{x} = \sqrt{7}$, then what is the value of $(x^2 + 1) ÷ [ x^4 +(\frac{1}{x^2})]$?

$\frac{1}{4}$

If $ x + \frac{1}{x} = \sqrt{7}$,

then what is the value of $(x^2 + 1) ÷ [ x^4 +(\frac{1}{x^2})]$

We can write $(x^2 + 1) ÷ [ x^4 +(\frac{1}{x^2})]$ this equation as (x + \(\frac{1}{x}\)) ÷ [ x³ + \(\frac{1}{x^3}\)] by dividing x on both numerator and denominator.

We know that,

If x + \(\frac{1}{x}\)  = n

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

 $x^3 +\frac{1}{x^3}$ = ($\sqrt{7}$)³ - 3 × $\sqrt{7}$ = $4\sqrt{7}$

Put these values in the required equation,

= $(x + \frac{1}{x})$ ÷ [ $x^3 +(\frac{1}{x^3})]$ = $\sqrt{7}$ ÷ $4\sqrt{7}$ = $\frac{1}{4}$