If $$x^{3}+\frac{3}{x}=4(a^{3}+b^{3})$$ and $$3x+\frac{1}{x^3}=4(a^{3}-b^{3})$$, then $$a^{2}-b^{2}$$ is equal to

Given : $$x^{3}+\frac{3}{x}=4(a^{3}+b^{3})$$ and $$3x+\frac{1}{x^3}=4(a^{3}-b^{3})$$

Adding the above equations, we get :

=> $$x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = 8a^3$$

=> $$(x + \frac{1}{x})^3 = (2a)^3$$

=> $$x + \frac{1}{x} = 2a$$ ------------------Eqn(1)

Similarly, subtracting the above equations, we get :

=> $$x - \frac{1}{x} = 2b$$ ---------------Eqn(2)

Now, adding eqns(1) & (2)

=> $$2(a + b) = 2x$$

=> $$(a + b) = x$$

SUbtracting eqn(2) from (1)

=> $$2(a - b) = \frac{2}{x}$$

=> $$(a - b) = \frac{1}{x}$$

To find : $$a^{2}-b^{2}$$

= $$(a + b) (a - b)$$

= $$(x) (\frac{1}{x}) = 1$$