If (x² + \(\frac{1}{x^2}\) - a)² + (x + \(\frac{1}{x}\) - b)² = 0, where a and b are real numbers and x ≠ 0, then find the value of a in terms of b.

b² - 2

(x² + \(\frac{1}{x^2}\) - a)² + (x + \(\frac{1}{x}\) - b)² = 0

⇒ (x² + \(\frac{1}{x^2}\) - a)² = 0

⇒  x² + \(\frac{1}{x^2}\) = a .... (i)

and 

⇒ (x + \(\frac{1}{x}\) - b)² = 0

⇒ x + \(\frac{1}{x}\) = b

squaring both sides →

x² + \(\frac{1}{x^2}\) + 2 = b²

a + 2 = b² (using ... (i))

a = b² - 2