Question 1209815
Let $x = 2^{1/3}$. Then $4^{1/3} = (2^2)^{1/3} = (2^{1/3})^2 = x^2$.
So, $a = x^2 + x + 1$.
We want to find the value of $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
Let's find a common denominator:
$$ \frac{3a^2 + 3a + 1}{a^3} $$
We know that $a = x^2 + x + 1$. Multiply both sides by $x-1$:
$$ a(x-1) = (x^2+x+1)(x-1) = x^3 - 1 $$
Since $x = 2^{1/3}$, we have $x^3 = 2$.
So, $a(x-1) = 2 - 1 = 1$.
Therefore, $x-1 = \frac{1}{a}$, which means $x = 1 + \frac{1}{a}$.
Now, $x^3 = 2$, so
$$ \left(1 + \frac{1}{a}\right)^3 = 2 $$
$$ 1 + \frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3} = 2 $$
$$ \frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3} = 2 - 1 $$
$$ \frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3} = 1 $$

Thus, the value of $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$ is 1.

Final Answer: The final answer is $\boxed{1}$