$a = 1 + {2^{1/3}} + {4^{1/3}}$, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = ?$

 We have, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} + {1^3} = {\left( {\frac{1}{a} + 1} \right)^3}$

So, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = {\left( {\frac{1}{a} + 1} \right)^3} - 1$

Now, $a = 1 + {2^{1/3}} + {4^{1/3}} = 1 + {2^{1/3}} + {({2^{1/3}})^2}$

Also, $({2^{1/3}} - 1)\left[ {{{({2^{1/3}})}^2} + {2^{1/3}} \times 1 + {1^2}} \right] = \left[ {{{({2^{1/3}})}^3} - {1^3}} \right] = 1$

So, $a = 1 + {2^{1/3}} + {4^{1/3}} = \frac{1}{{{2^{1/3}} - 1}}$

Thus, $\frac{1}{a} = {2^{1/3}} - 1$

Or, $\frac{1}{a} + 1 = {2^{1/3}}$

Or, ${\left( {\frac{1}{a} + 1} \right)^3} = 2$

Thus, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = {\left( {\frac{1}{a} + 1} \right)^3} - 1 = 2 - 1 = 1$