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?
Answer & explanation
Correct answer: option 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