$$\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}} = x$$, then the value of $$\frac{3bx^2+3b}{ax}$$ is

We are given: $$x=\dfrac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$$

Rationalising the RHS

$$x=\dfrac{(\sqrt{a+3b}+\sqrt{a-3b})^2}{(\sqrt{a+3b}-\sqrt{a-3b})(\sqrt{a+3b}+\sqrt{a-3b})}$$

$$x=\dfrac{a+3b+a-3b+2\sqrt{a^2-9b^2}}{a+3b-a+3b}$$

$$x=\dfrac{a+\sqrt{a^2-9b^2}}{3b}$$ ....(1)

Now, squaring both sides

$$x^2=\dfrac{a^2+a^2-9b^2+2a\sqrt{a^2-9b^2}}{9b^2}$$

$$x^2=\dfrac{2a^2+2a\sqrt{a^2-9b^2}-9b^2}{9b^2}$$

$$x^2=\dfrac{2a^2+2a\sqrt{a^2-9b^2}}{9b^2}-1$$

$$x^2+1=\dfrac{2a^2+2a\sqrt{a^2-9b^2}}{9b^2}$$

$$3b(x^2+1)=\dfrac{2a^2+2a\sqrt{a^2-9b^2}}{3b}$$

$$\dfrac{3b(x^2+1)}{a}=\dfrac{2a+2\sqrt{a^2-9b^2}}{3b}$$

$$\dfrac{3b(x^2+1)}{a}=2\times\left(\dfrac{a+\sqrt{a^2-9b^2}}{3b}\right)$$

$$\dfrac{3b(x^2+1)}{a}=2\times x$$ [from 1]

$$\dfrac{3b(x^2+1)}{ax}=2$$

$$\dfrac{3bx^2+3b}{ax}=2$$