The solution of $$\frac{\sqrt[7]{12+x}}{x} + \frac{\sqrt[7]{12+x}}{12} = \frac{64}{3}\left(\sqrt[7]{x}\right)$$ is of the form $$\frac{a}{b}$$ where $$a, b$$ are natural numbers with $$\text{GCD}(a,b) = 1$$; then $$(b-a)$$ is equal to

$$\dfrac{\sqrt[7]{12+x}}{x} + \dfrac{\sqrt[7]{12+x}}{12} = \dfrac{64}{3}\left(\sqrt[7]{x}\right)$$

Taking $$\sqrt[7]{12+x}$$ as common we get

$$\sqrt[7]{12+x} \left( \dfrac{1}{x} + \dfrac{1}{12}\right) = \dfrac{64}{3}\left(\sqrt[7]{x}\right)$$

$$\sqrt[7]{12+x} \left( \dfrac{x+12}{12x}\right) = \dfrac{64}{3}\left(\sqrt[7]{x}\right)$$

Transposing $$12x$$ to the other side of the equation we get

$$\sqrt[7]{12+x} (12+x) = \dfrac{64}{3}\left(12x\sqrt[7]{x}\right)$$

Now, we can $$\sqrt[7]{12+x}$$ as $${(12+x)}^{\frac{1}{7}}$$. Using the exponent rule that $$x^a.x^b = x^{a+b}$$, we can write

$${(12+x)}^{1+\frac{1}{7}} = (64 \times \dfrac{12}{3}) x^{1+\frac{1}{7}}$$

Simplifying the terms we get

$${(12+x)}^{\frac{8}{7}} =( 64 \times 4) x^{\frac{8}{7}}$$

$$64$$ can be expressed as $$2^6$$ and $$4 = 2^2$$. Thus, $$64 \times 4 = 2^8$$

$${(12+x)}^{\frac{8}{7}} = 2^8 x^{\frac{8}{7}}$$

Taking the eighth root we get

$${(12+x)}^{\frac{1}{7}} = 2 x^{\frac{1}{7}}$$

And raising the expression to the power of 7 we get

$$12+x = 128x \implies 127x = 12 \ \text{or}\ x = \dfrac{12}{127} $$

We can observe that $$12,127$$ have no common factors hence their $$gcd = 1$$

Thus $$ b-a = 127-12 = 115$$