1. **Problem statement:** Given the equation $a^3 + 1 - \sqrt{6}a = 0$, find the value of $a^3 + \frac{a^3}{3}$. 2. **Rewrite the problem:** We want to find $$a^3 + \frac{a^3}{3} = a^3 \left(1 + \frac{1}{3}\right) = \frac{4}{3}a^3.$$ 3. **From the given equation:** $$a^3 + 1 - \sqrt{6}a = 0 \implies a^3 = \sqrt{6}a - 1.$$ 4. **Substitute $a^3$ into the expression:** $$\frac{4}{3}a^3 = \frac{4}{3}(\sqrt{6}a - 1) = \frac{4}{3}\sqrt{6}a - \frac{4}{3}.$$ 5. **Find $a$ to evaluate the expression:** Rewrite the original equation as $$a^3 - \sqrt{6}a + 1 = 0.$$ 6. **Try to find roots:** Let us check if $a = \sqrt{2}$ is a root: $$a^3 = (\sqrt{2})^3 = 2\sqrt{2},$$ $$\sqrt{6}a = \sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3}.$$ Substitute: $$2\sqrt{2} + 1 - 2\sqrt{3} \neq 0,$$ so $a=\sqrt{2}$ is not a root. 7. **Try $a = \sqrt{3}$:** $$a^3 = (\sqrt{3})^3 = 3\sqrt{3},$$ $$\sqrt{6}a = \sqrt{6} \times \sqrt{3} = \sqrt{18} = 3\sqrt{2}.$$ Substitute: $$3\sqrt{3} + 1 - 3\sqrt{2} \neq 0,$$ so $a=\sqrt{3}$ is not a root. 8. **Try $a = 1$:** $$1^3 + 1 - \sqrt{6} \times 1 = 1 + 1 - \sqrt{6} = 2 - \sqrt{6} \neq 0.$$ 9. **Try $a = -1$:** $$(-1)^3 + 1 - \sqrt{6} \times (-1) = -1 + 1 + \sqrt{6} = \sqrt{6} \neq 0.$$ 10. **Try $a = \sqrt{6}/3$:** $$a^3 = \left(\frac{\sqrt{6}}{3}\right)^3 = \frac{(\sqrt{6})^3}{27} = \frac{6\sqrt{6}}{27} = \frac{2\sqrt{6}}{9},$$ $$\sqrt{6}a = \sqrt{6} \times \frac{\sqrt{6}}{3} = \frac{6}{3} = 2.$$ Substitute: $$a^3 + 1 - \sqrt{6}a = \frac{2\sqrt{6}}{9} + 1 - 2 = \frac{2\sqrt{6}}{9} - 1 \neq 0.$$ 11. **Since exact root is complicated, use the original equation to express $a^3$ and leave answer in terms of $a$: $$\boxed{\frac{4}{3}a^3 = \frac{4}{3}(\sqrt{6}a - 1) = \frac{4}{3}\sqrt{6}a - \frac{4}{3}}.$$ This is the simplified exact form of the value requested.