If x^2-4 x+1=0, then the value of 16≤ft(x^4-1/(x^4)) is

If $x^2-4 x+1=0$

Then the value of $16\left(x^4-\frac{1}{x^4}\right)$

Divide by x on both sides of $x^2-4 x+4=0$

x + \(\frac{4}{x}\) = 4

Put the value of x = 2 (This value will satisfy the equation)

$16\left(x^4-\frac{1}{x^4}\right)$ = $16\left(2^4-\frac{1}{2^4}\right)$

$16\left(x^4-\frac{1}{x^4}\right)$ = 16(\(\frac{256 - 1}{16}\))

$16\left(x^4-\frac{1}{x^4}\right)$ = 16 × \(\frac{255}{16}\) = 255