If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6-5 x^3-1}{x^6+7 x^3-1} ?$

$\frac{45}{49}$

If $x-\frac{1}{x}=5, x \neq 0$

$\frac{x^6-5 x^3-1}{x^6+7 x^3-1} $

we can write $\frac{x^6-5 x^3-1}{x^6+7 x^3-1} $ as $\frac{x^3 -\frac{1}{x^3}-5}{x^3 -\frac{1}{x^3} + 7} $ ----(A)

If x - \(\frac{1}{x}\)  = n

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

then, $x^3 - \frac{1}{x^3}$ = 5³ + 3 × 5 = 140

Put the value of $x^3 - \frac{1}{x^3}$ in eq. (A)

$\frac{x^3 -\frac{1}{x^3}-5}{x^3 -\frac{1}{x^3} + 7} $ = \(\frac{140 - 5}{140 + 7}\) = $\frac{45}{49}$