1. **State the problem:** Find the maximum and minimum values of the function $f(x,y,z) = xyz$ subject to the constraint $g(x,y,z) = xy + xz + yz = 108$. 2. **Method:** Use Lagrange multipliers. We introduce a multiplier $\lambda$ and solve the system: $$\nabla f = \lambda \nabla g$$ with the constraint $g(x,y,z) = 108$. 3. **Calculate gradients:** $$\nabla f = \left( yz, xz, xy \right)$$ $$\nabla g = \left( y+z, x+z, x+y \right)$$ 4. **Set up equations:** $$yz = \lambda (y+z)$$ $$xz = \lambda (x+z)$$ $$xy = \lambda (x+y)$$ 5. **Analyze the system:** Assume $x,y,z \neq 0$ (otherwise $f=0$ which is a candidate minimum). Divide each equation by $xyz$: $$\frac{1}{x} = \lambda \left( \frac{1}{x} + \frac{1}{y} \right), \quad \frac{1}{y} = \lambda \left( \frac{1}{x} + \frac{1}{z} \right), \quad \frac{1}{z} = \lambda \left( \frac{1}{y} + \frac{1}{z} \right)$$ 6. **Symmetry suggests $x=y=z$:** Let $x=y=z=t$. Then constraint: $$3t^2 = 108 \implies t^2 = 36 \implies t = \pm 6$$ 7. **Evaluate $f$ at $x=y=z=6$:** $$f = 6 \times 6 \times 6 = 216$$ At $x=y=z=-6$: $$f = (-6) \times (-6) \times (-6) = -216$$ 8. **Check boundary cases:** If any variable is zero, $f=0$ which lies between $-216$ and $216$. 9. **Conclusion:** - Maximum value of $f$ is $216$ at $(6,6,6)$. - Minimum value of $f$ is $-216$ at $(-6,-6,-6)$. Hence, the extrema are: $$\boxed{f_{max} = 216, \quad f_{min} = -216}$$