If ≤ft(2 x-3/x)=2, then what is the value of ≤ft(16 x^4+81/(x^4)) ?

If $(x - \frac{1}{x}) = n$

(x² + x^-2) = (n)² + 2 = b

= (x⁴ + x^-4) = b² - 2

According to the question,

$\left(2 x-\frac{3}{x}\right)=2$

$\left(4 x^2+\frac{9}{x^2}\right) $ = (2)² + 2 × 2 × 3 = 16

$\left(16 x^4+\frac{81}{x^4}\right)$ = 16² - 2 × 2 × 3 × 2 × 3 

$\left(16 x^4+\frac{81}{x^4}\right)$ = 256 - 72 = 184