Let `a^3+b^2=4`. In the expansion of `(ax^(1/9)+bx^(-1/6))^(10)`, the term independent of x is 10k. Find the maximum value of k.

Answer

To solve the problem, we need to find the maximum value of \( k \) given the equation \( a^3 + b^2 = 4 \) and the expression \( (ax^{1/9} + bx^{-1/6})^{10} \). We are looking for the term independent of \( x \) in the expansion of this expression.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the expansion of \( (ax^{1/9} + bx^{-1/6})^{10} \) can be expressed as:
\[ T_r = \binom{10}{r} (a)^{r} (b)^{10-r} x^{\frac{r}{9} - \frac{10-r}{6}} \]

2. **Simplify the Exponent of \( x \)**:
We need to simplify the exponent of \( x \):
\[ \frac{r}{9} - \frac{10 - r}{6} = \frac{r}{9} - \frac{10}{6} + \frac{r}{6} = \frac{r}{9} + \frac{r}{6} - \frac{5}{3} \]
To combine the fractions, find a common denominator (which is 18):
\[ = \frac{2r}{18} + \frac{3r}{18} - \frac{30}{18} = \frac{5r - 30}{18} \]

3. **Set the Exponent to Zero**:
For the term to be independent of \( x \), we set the exponent equal to zero:
\[ 5r - 30 = 0 \implies 5r = 30 \implies r = 6 \]

4. **Substitute \( r = 6 \) into the General Term**:
Now substitute \( r = 6 \) into the general term:
\[ T_6 = \binom{10}{6} a^6 b^4 \]

5. **Calculate the Binomial Coefficient**:
The binomial coefficient \( \binom{10}{6} \) can be calculated as:
\[ \binom{10}{6} = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
Thus, we have:
\[ T_6 = 210 a^6 b^4 \]

6. **Relate to Given Condition**:
From the problem, we know:
\[ 210 a^6 b^4 = 10k \implies k = 21 a^6 b^4 \]

7. **Using the Condition \( a^3 + b^2 = 4 \)**:
We apply the AM-GM inequality:
\[ \frac{a^3 + b^2}{2} \geq \sqrt{a^3 b^2} \]
Given \( a^3 + b^2 = 4 \):
\[ 2 \geq \sqrt{a^3 b^2} \implies 4 \geq a^3 b^2 \]
Thus, we have:
\[ a^3 b^2 \leq 4 \]

8. **Express \( a^6 b^4 \)**:
We can express \( a^6 b^4 \) in terms of \( a^3 b^2 \):
\[ a^6 b^4 = (a^3 b^2)^2 \leq 4^2 = 16 \]

9. **Substituting Back to Find Maximum \( k \)**:
Therefore, substituting back:
\[ k = 21 a^6 b^4 \leq 21 \times 16 = 336 \]

### Conclusion:
The maximum value of \( k \) is:
\[ \boxed{336} \]