1. We need to find the greatest term of the sequence $a_n=\frac{n^3}{n^4+147}, \quad n=1,2,3,\dots$ and determine the corresponding index $\alpha$ such that $a_\alpha$ is maximum.

2. Since the sequence is defined on positive integers, we analyze the function $f(x)=\frac{x^3}{x^4+147}, \quad x>0$ and then check nearby integers.

3. Differentiate using the quotient rule: $f'(x)=\frac{3x^2(x^4+147)-x^3(4x^3)}{(x^4+147)^2}$ $=\frac{3x^6+441x^2-4x^6}{(x^4+147)^2}$ $=\frac{-x^6+441x^2}{(x^4+147)^2}$ $=\frac{x^2(441-x^4)}{(x^4+147)^2}$

4. Since the denominator is always positive and $x^2>0$ for $x>0$, the sign of $f'(x)$ depends on $441-x^4$ So:

   • $f'(x)>0$ when $x^4<441$
   • $f'(x)<0$ when $x^4>441$

5. Now, $441=21^2$ and $\sqrt{441}=21$ Hence $x^4=441 \implies x=\sqrt{21}$ Therefore:

   • $f(x)$ increases for $x<\sqrt{21}$
   • $f(x)$ decreases for $x>\sqrt{21}$

   So the maximum for integer $n$ must occur near $n=\sqrt{21}\approx 4.58$.

6. Check the nearby integers $n=4$ and $n=5$: $a_4=\frac{4^3}{4^4+147}=\frac{64}{256+147}=\frac{64}{403}$ $a_5=\frac{5^3}{5^4+147}=\frac{125}{625+147}=\frac{125}{772}$

7. Compare them: $\frac{64}{403} \quad \text{and} \quad \frac{125}{772}$ Cross-multiply: $64\cdot 772=49408$ $125\cdot 403=50375$ Since $50375>49408,$ we get $\frac{125}{772}>\frac{64}{403}$ Hence, $a_5>a_4$

8. Since the function increases up to about $4.58$ and decreases after that, the greatest term in the sequence occurs at $n=5$

Therefore, $\alpha=5$