If $${x^3} + \frac{3}{x}$$   = $$4\left( {{a^3} + {b^3}} \right)$$   and $$3x + \frac{1}{{{x^3}}}$$   = $$4\left( {{a^3} - {b^3}} \right){\text{,}}$$   then a² - b² is equal to?

Solution (By Examveda Team)

$$\eqalign{ & {x^3} + \frac{3}{x} = 4\left( {{a^3} + {b^3}} \right)\, . . . . . {\text{(i)}} \cr & 3x + \frac{1}{{{x^3}}} = 4\left( {{a^3} - {b^3}} \right)\, . . . . . (ii) \cr & {\text{Equation (i)}} + {\text{(ii)}} \cr & {\left( {x + \frac{1}{x}} \right)^3} = 8{a^3} \cr & \Leftrightarrow x + \frac{1}{x} = 2a\, . . . . . (iii) \cr & Equation{\text{ }}(i) - (ii) \cr & x - \frac{1}{x} = 2b\, . . . . . (iv) \cr & Equation{\text{ }}(iii) - (iv) \cr & 2\left( {a - b} \right) = \frac{2}{x} \cr & \Rightarrow a - b = \frac{1}{x} \cr & \Rightarrow a + b = x \cr & \because {a^2} - {b^2} = \left( {a + b} \right) \times \left( {a - b} \right) \cr & \Rightarrow {a^2} - {b^2} = x \times \frac{1}{x} \cr & \Rightarrow {a^2} - {b^2} = 1 \cr} $$