If a, b, c are non-zero and ${14}^a={36}^b={84}^c$, then 6b[ $\mathrm{}\frac{1}{c}-\frac{1}{a}$ ]is equal to

The correct answer is 3.

We are given that a, b, c are non-zero and ${14}^a={36}^b={84}^c$.

Step 1 — Introduce a common variable.
Let ${14}^a={36}^b={84}^c=k$.

From this we can write:
$14=k^{1/a}$
$36=k^{1/b}$
$84=k^{1/c}$

Step 2 — Use the key numerical relationship 84 ÷ 14 = 6.
Dividing the expression for 84 by the expression for 14:

$\frac{84}{14}=\frac{k^{1/c}}{k^{1/a}}$

$6=k^{(1/c-1/a)}$  … (i)

Step 3 — Use the key numerical relationship 36 = 6².
Since $36=6^2$, we write:

$k^{1/b}=6^2$

Raising both sides to the power $\frac{1}{2}$:

$6=k^{1/\left(2b\right)}$  … (ii)

Step 4 — Equate equations (i) and (ii).
Both expressions equal 6, so their exponents of k must be equal:

$\frac{1}{c}-\frac{1}{a}=\frac{1}{2b}$

Step 5 — Multiply both sides by 6b.

$6b(\frac{1}{c}-\frac{1}{a})=6b\times \frac{1}{2b}$

$6b(\frac{1}{c}-\frac{1}{a})=\frac{6}{2}=3$

Therefore, $6b(\frac{1}{c}-\frac{1}{a})=3$.