If (x³ + \(\frac{1}{x^3}\) - a)² + (x + \(\frac{1}{x}\) - b)² = 0, where a and b are real numbers and x ≠ 0, then \(\frac{a}{b}\) is equal to?

b² - 3

(x³ + \(\frac{1}{x^3}\) - a)² + (x + \(\frac{1}{x}\) - b)² = 0

⇒ (x³ + \(\frac{1}{x^3}\) - a)² = 0

⇒ x³ + \(\frac{1}{x^3}\) = a .... (i)

and 

⇒ (x + \(\frac{1}{x}\) - b)² = 0

⇒ x + \(\frac{1}{x}\) = b

cubing both sides →

x³ + \(\frac{1}{x^3}\) + 3x × \(\frac{1}{x}\) (b) = b³

a + 3b = b³ (using ... (i))

a = b³ - 3b

Now, \(\frac{a}{b}\) = \(\frac{b^3 - 3b}{b}\) = b² - 3