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Step-by-Step Solution
Step 1: Write the general term of the binomial expansion
Consider the expansion of
$ ( \alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}} )^{10} $.
The general term $ T_{r+1} $ (the term containing $ x $ raised to some power) is:
$$
T_{r+1} = \binom{10}{r} \alpha^{10 - r} \left(x^{\frac{1}{9}}\right)^{10 - r} \; \beta^r \left(x^{-\frac{1}{6}}\right)^r.
$$
Simplifying the exponent of $ x $:
$$
T_{r+1}
= \binom{10}{r} \alpha^{10-r} \beta^r \; x^{\frac{10-r}{9} - \frac{r}{6}}.
$$
Step 2: Find the condition for the term to be independent of x
For the term to be independent of $ x $, the overall exponent of $ x $ must be zero:
$$
\frac{10-r}{9} - \frac{r}{6} = 0.
$$
Solve for $ r $:
$$
\frac{10-r}{9} = \frac{r}{6}
\quad \Longrightarrow \quad
60 - 6r = 9r
\quad \Longrightarrow \quad
60 = 15r
\quad \Longrightarrow \quad
r = 4.
$$
Step 3: Determine the independent term
Substituting $ r = 4 $ into the expression for $ T_{r+1} $:
$$
T_{5} = \binom{10}{4} \alpha^{10 - 4} \beta^4
= \binom{10}{4} \alpha^6 \beta^4.
$$
Step 4: Use the given condition $ \alpha^3 + \beta^2 = 4 $ and AM-GM inequality
We have $ \alpha^3 + \beta^2 = 4 $, with $ \alpha > 0 $ and $ \beta > 0 $.
From the AM-GM inequality on $ \alpha^3 $ and $ \beta^2 $:
$$
\frac{\alpha^3 + \beta^2}{2}
\;\ge\;
\sqrt{\alpha^3 \, \beta^2}.
$$
Since $ \alpha^3 + \beta^2 = 4 $, we get
$$
\frac{4}{2} = 2
\;\ge\;
\sqrt{\alpha^3 \beta^2}
\quad \Longrightarrow \quad
4 \;\ge\; \alpha^3 \beta^2.
$$
Hence,
$$
\alpha^6 \beta^4 \;\le\; ( \alpha^3 \beta^2 )^2 \;\le\; 16.
$$
Step 5: Find the maximum value of the independent term
Thus, the term $ \alpha^6 \beta^4 $ can be at most 16. Therefore, the maximum possible value of the independent term is:
$$
T_{5, \max} = \binom{10}{4} \times 16.
$$
Note that $ \binom{10}{4} = 210 $. So,
$$
T_{5, \max} = 210 \times 16 = 3360.
$$
According to the question, the maximum value of the independent term is $ 10k $. That is,
$$
10k = 3360
\quad \Longrightarrow \quad
k = 336.
$$
Step 6: Conclude the correct answer
Thus, the required value of $ k $ is
336.
