Problem 1833

1. Let $s = \sqrt[3]{\frac{a+1}{2} + \frac{a+3}{6}\sqrt{\frac{4a+3}{3}}}$ and $t = \sqrt[3]{\frac{a+1}{2} - \frac{a+3}{6}\sqrt{\frac{4a+3}{3}}}$. We need to prove that $s + t$ is independent of $a$ for $a \ge -\frac{3}{4}$.

2. Consider the identity for the sum of cubes:
$$(s + t)^3 = s^3 + t^3 + 3st(s + t)$$
Here, $s^3$ and $t^3$ are given by:
$$s^3 = \frac{a+1}{2} + \frac{a+3}{6}\sqrt{\frac{4a+3}{3}}$$
$$t^3 = \frac{a+1}{2} - \frac{a+3}{6}\sqrt{\frac{4a+3}{3}}$$

3. Adding these two expressions, we get:
$$s^3 + t^3 = \left( \frac{a+1}{2} + \frac{a+3}{6}\sqrt{\frac{4a+3}{3}} \right) + \left( \frac{a+1}{2} - \frac{a+3}{6}\sqrt{\frac{4a+3}{3}} \right) = a + 1$$

4. Next, we need to find $st$. We have:
$$s^3 t^3 = \left( \frac{a+1}{2} \right)^2 - \left( \frac{a+3}{6} \right)^2 \left( \frac{4a+3}{3} \right)$$
Simplifying the right-hand side:
$$s^3 t^3 = \frac{(a+1)^2}{4} - \frac{(a+3)^2 (4a+3)}{108}$$
$$= \frac{a^2 + 2a + 1}{4} - \frac{(a^2 + 6a + 9)(4a + 3)}{108}$$
$$= \frac{a^2 + 2a + 1}{4} - \frac{4a^3 + 27a^2 + 54a + 27}{108}$$
$$= \frac{a^2 + 2a + 1}{4} - \frac{4a^3 + 27a^2 + 54a + 27}{108}$$
$$= \frac{27(a^2 + 2a + 1) - (4a^3 + 27a^2 + 54a + 27)}{108}$$
$$= \frac{27a^2 + 54a + 27 - 4a^3 - 27a^2 - 54a - 27}{108}$$
$$= -\frac{4a^3}{108} = -\frac{a^3}{27}$$
Therefore:
$$s^3 t^3 = -\frac{a^3}{27} \implies (st)^3 = -\frac{a^3}{27} \implies st = -\frac{a}{3}$$

5. Using the identity $(s + t)^3 = s^3 + t^3 + 3st(s + t)$, we substitute the known values:
$$(s + t)^3 = (a + 1) + 3\left(-\frac{a}{3}\right)(s + t)$$
$$(s + t)^3 = a + 1 - a(s + t)$$
Let $y = s + t$:
$$y^3 = a + 1 - ay$$
$$y^3 + ay - (a + 1) = 0$$

6. Notice that the coefficients of the polynomial $y^3 + ay - (a + 1) = 0$ sum to zero, which implies $y = 1$ is a root:
$$1^3 + a \cdot 1 - (a + 1) = 1 + a - a - 1 = 0$$

7. Since $y = 1$ is a solution, we factor the polynomial:
$$y^3 + ay - (a + 1) = (y - 1)(y^2 + y + a + 1)$$

8. For $a \ge -\frac{3}{4}$, the other roots of the quadratic $y^2 + y + a + 1 = 0$ are complex, but the expression $s + t$ is real-valued. Therefore, the only real solution is $y = 1$.

9. For $a = -\frac{3}{4}$, we can check directly:
$$s + t = \sqrt[3]{\frac{-\frac{3}{4} + 1}{2} + \frac{-\frac{3}{4} + 3}{6}\sqrt{\frac{4(-\frac{3}{4}) + 3}{3}}} + \sqrt[3]{\frac{-\frac{3}{4} + 1}{2} - \frac{-\frac{3}{4} + 3}{6}\sqrt{\frac{4(-\frac{3}{4}) + 3}{3}}}$$
Simplifying:
$$s + t = \sqrt[3]{\frac{1}{8}} + \sqrt[3]{\frac{1}{8}} = \frac{1}{2} + \frac{1}{2} = 1$$

Thus, the sum $\sqrt[3]{\frac{a+1}{2} + \frac{a+3}{6}\sqrt{\frac{4a+3}{3}}} + \sqrt[3]{\frac{a+1}{2} - \frac{a+3}{6}\sqrt{\frac{4a+3}{3}}}$ is indeed independent of $a$ and equals 1.

The final answer is $\boxed{ 1 }$