1. Given: $AB=3$, $BC=\sqrt{3}$, $AC=b$, and $\angle ACD=150^\circ$. We need to find the value of $4b^2 - 6\sqrt{3}$. 2. Since $A$, $B$, and $C$ form a triangle, we can use the Law of Cosines on $\triangle ABC$ to relate $b$ to the other sides and angle. 3. However, the problem gives $\angle ACD=150^\circ$, which involves point $D$ on the extension of $C$. We need to interpret this carefully. 4. Let's consider vectors or coordinate geometry to express $b$ in terms of given lengths and angles. 5. Place point $C$ at the origin $(0,0)$ for convenience. 6. Let $B$ be at $(x_B,y_B)$ such that $CB=\sqrt{3}$. Since $BC=\sqrt{3}$, place $B$ at $(\sqrt{3},0)$ on the positive x-axis. 7. The angle $\angle ACD=150^\circ$ means the angle between vectors $CA$ and $CD$ is $150^\circ$. Since $D$ lies on the extension from $C$, we can place $D$ at some point on the line making $150^\circ$ with $CA$. 8. Let vector $CA$ make an angle $\theta$ with the positive x-axis. Then vector $CD$ makes an angle $\theta + 150^\circ$. 9. Since $B$ is at $(\sqrt{3},0)$ and $C$ at $(0,0)$, vector $CB$ points along the positive x-axis. 10. Given $AB=3$, we can write coordinates of $A$ as $(b\cos\theta, b\sin\theta)$. 11. Distance $AB=3$ gives: $$AB^2 = (b\cos\theta - \sqrt{3})^2 + (b\sin\theta - 0)^2 = 9$$ 12. Expanding: $$ (b\cos\theta - \sqrt{3})^2 + (b\sin\theta)^2 = 9 $$ $$ b^2\cos^2\theta - 2b\sqrt{3}\cos\theta + 3 + b^2\sin^2\theta = 9 $$ 13. Using $\cos^2\theta + \sin^2\theta = 1$: $$ b^2 - 2b\sqrt{3}\cos\theta + 3 = 9 $$ 14. Simplify: $$ b^2 - 2b\sqrt{3}\cos\theta = 6 $$ 15. Rearranged: $$ b^2 = 6 + 2b\sqrt{3}\cos\theta $$ 16. Now, consider $\angle ACD=150^\circ$. Since $CD$ makes an angle $\theta + 150^\circ$ and $CB$ is along $0^\circ$, the angle between $CB$ and $CD$ is $150^\circ$. 17. The vector $CD$ is at $\theta + 150^\circ$, and $CB$ is at $0^\circ$, so: $$ \theta + 150^\circ = 0^\circ + 150^\circ \implies \theta = 0^\circ $$ 18. So $\theta=0^\circ$, meaning $A$ lies on the positive x-axis. 19. Substitute $\cos 0^\circ = 1$ into step 14: $$ b^2 - 2b\sqrt{3} = 6 $$ 20. Rearranged: $$ b^2 = 6 + 2b\sqrt{3} $$ 21. We want to find $4b^2 - 6\sqrt{3}$. 22. Multiply both sides of step 20 by 4: $$ 4b^2 = 24 + 8b\sqrt{3} $$ 23. So: $$ 4b^2 - 6\sqrt{3} = 24 + 8b\sqrt{3} - 6\sqrt{3} = 24 + (8b - 6)\sqrt{3} $$ 24. To find the numeric value, solve for $b$ from step 19: $$ b^2 - 2b\sqrt{3} - 6 = 0 $$ 25. Use quadratic formula for $b$: $$ b = \frac{2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 + 4 \times 6}}{2} = \frac{2\sqrt{3} \pm \sqrt{12 + 24}}{2} = \frac{2\sqrt{3} \pm \sqrt{36}}{2} $$ 26. Simplify: $$ b = \frac{2\sqrt{3} \pm 6}{2} = \sqrt{3} \pm 3 $$ 27. Since $b$ is a length, it must be positive. Both $\sqrt{3} + 3$ and $3 - \sqrt{3}$ are positive, but $3 - \sqrt{3}$ is smaller and more reasonable for a triangle side. 28. Take $b = 3 - \sqrt{3}$. 29. Calculate $8b - 6$: $$ 8b - 6 = 8(3 - \sqrt{3}) - 6 = 24 - 8\sqrt{3} - 6 = 18 - 8\sqrt{3} $$ 30. Substitute back into step 23: $$ 4b^2 - 6\sqrt{3} = 24 + (18 - 8\sqrt{3})\sqrt{3} = 24 + 18\sqrt{3} - 8 \times 3 = 24 + 18\sqrt{3} - 24 = 18\sqrt{3} $$ 31. Final answer: $$ \boxed{18\sqrt{3}} $$