Exponents and powers relation

Let $14^a = 36^b = 84^c = k$. Then $14 = k^{1/a}$, $36 = k^{1/b} \implies 6 = k^{1/(2b)}$, $84 = k^{1/c}$.

We know $84 = 14 \times 6$. Substituting the powers of $k$: $k^{1/c} = k^{1/a} \times k^{1/(2b)} = k^{1/a + 1/(2b)}$. So, $\frac{1}{c} = \frac{1}{a} + \frac{1}{2b} \implies \frac{1}{c} - \frac{1}{a} = \frac{1}{2b}$.

Thus, $6b \left(\frac{1}{c} - \frac{1}{a}\right) = 6b \times \frac{1}{2b} = 3$.