1. Let the initial ratio of $\dfrac{^{14}\mathrm{C}}{^{12}\mathrm{C}}$ in the living wood be the same as that in the atmosphere.

2. After the wood is cut, $^{14}\mathrm{C}$ starts decaying while $^{12}\mathrm{C}$ remains unchanged. Hence, the ratio $\dfrac{^{14}\mathrm{C}}{^{12}\mathrm{C}}$ decreases exactly like the amount of $^{14}\mathrm{C}$.

3. Given: $\frac{\left(\dfrac{^{14}\mathrm{C}}{^{12}\mathrm{C}}\right)_{\text{wood}}}{\left(\dfrac{^{14}\mathrm{C}}{^{12}\mathrm{C}}\right)_{\text{atmosphere}}} = \frac{1}{8}$ So, the amount of $^{14}\mathrm{C}$ left is: $\frac{N}{N_0} = \frac{1}{8}$

4. Radioactive decay formula: $N = N_0\left(\frac{1}{2}\right)^{t/T_{1/2}}$ where $T_{1/2} = 5730$ years.

5. Substitute the given ratio: $\frac{1}{8} = \left(\frac{1}{2}\right)^{t/5730}$

6. Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$ we get: $\left(\frac{1}{2}\right)^{t/5730} = \left(\frac{1}{2}\right)^3$

7. Equating powers: $\frac{t}{5730} = 3$ $t = 3 \times 5730 = 17190$

8. Therefore, the age of the wood sample is: $\boxed{17190\text{ years}}$

Comparison with stored correct answer:

• Stored correct answer = $17190$
• Derived answer = $17190$
• They match.