1. **State the problem:** Given the expression $$(\sqrt{a})^2 - 2\sqrt{3}a^2 + 3a = 4a,$$ prove that $$\frac{a^3 + a^2 + 1}{3a\sqrt{6a} + a^2} = 1.$$\n\n2. **Simplify the given expression:** Note that $$(\sqrt{a})^2 = a.$$ So the left side becomes $$a - 2\sqrt{3}a^2 + 3a = 4a.$$\n\n3. **Combine like terms:** $$a + 3a - 2\sqrt{3}a^2 = 4a \implies 4a - 2\sqrt{3}a^2 = 4a.$$\n\n4. **Subtract $4a$ from both sides:** $$4a - 2\sqrt{3}a^2 - 4a = 0 \implies -2\sqrt{3}a^2 = 0.$$\n\n5. **Divide both sides by $-2a^2$ (assuming $a \neq 0$):** $$\sqrt{3} = 0,$$ which is false unless $a=0$. So the given equation holds only if $a=0$.\n\n6. **Check the expression to prove:** $$\frac{a^3 + a^2 + 1}{3a\sqrt{6a} + a^2} = 1.$$\n\n7. **Multiply both sides by the denominator:** $$a^3 + a^2 + 1 = 3a\sqrt{6a} + a^2.$$\n\n8. **Subtract $a^2$ from both sides:** $$a^3 + 1 = 3a\sqrt{6a}.$$\n\n9. **Rewrite $3a\sqrt{6a}$ as $3a \times \sqrt{6a} = 3a \times \sqrt{6} \times \sqrt{a} = 3\sqrt{6} a^{3/2}.$**\n\n10. **So the equation becomes:** $$a^3 + 1 = 3\sqrt{6} a^{3/2}.$$\n\n11. **Let $x = a^{3/2}$. Then $a^3 = (a^{3/2})^2 = x^2$. The equation becomes:** $$x^2 + 1 = 3\sqrt{6} x.$$\n\n12. **Rearranged as a quadratic in $x$:** $$x^2 - 3\sqrt{6} x + 1 = 0.$$\n\n13. **Solve using quadratic formula:** $$x = \frac{3\sqrt{6} \pm \sqrt{(3\sqrt{6})^2 - 4}}{2} = \frac{3\sqrt{6} \pm \sqrt{54 - 4}}{2} = \frac{3\sqrt{6} \pm \sqrt{50}}{2}.$$\n\n14. **Simplify $\sqrt{50} = 5\sqrt{2}$:** $$x = \frac{3\sqrt{6} \pm 5\sqrt{2}}{2}.$$\n\n15. **Recall $x = a^{3/2}$, so $a$ must be positive and real.**\n\n16. **Therefore, the original expression holds for $a$ such that $a^{3/2} = \frac{3\sqrt{6} \pm 5\sqrt{2}}{2}$.**\n\n**Final answer:** The given expression $$\frac{a^3 + a^2 + 1}{3a\sqrt{6a} + a^2} = 1$$ holds true for values of $a$ satisfying $$a^{3/2} = \frac{3\sqrt{6} \pm 5\sqrt{2}}{2}.$$