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 term independent of \( x \) in the expansion of \( (ax^{1/9} + bx^{-1/6})^{10} \).

### Step-by-Step Solution:

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

2. **Determine the Power of \( x \):**
The exponent of \( x \) in \( T_r \) is:
\[ \frac{10 - r}{9} - \frac{r}{6} \]
To combine these fractions, we find a common denominator (which is 18):
\[ \frac{10 - r}{9} = \frac{2(10 - r)}{18} = \frac{20 - 2r}{18} \]
\[ -\frac{r}{6} = -\frac{3r}{18} \]
Thus, the exponent of \( x \) becomes:
\[ \frac{20 - 2r - 3r}{18} = \frac{20 - 5r}{18} \]

3. **Find the Term Independent of \( x \):**
For the term to be independent of \( x \), we set the exponent equal to zero:
\[ 20 - 5r = 0 \implies r = 4 \]

4. **Substitute \( r \) into the General Term:**
Now, substituting \( r = 4 \) into the general term:
\[ T_4 = \binom{10}{4} a^{10-4} b^4 = \binom{10}{4} a^6 b^4 \]

5. **Calculate \( \binom{10}{4} \):**
\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
Therefore,
\[ T_4 = 210 a^6 b^4 \]

6. **Relate to \( 10k \):**
We know that this term equals \( 10k \):
\[ 210 a^6 b^4 = 10k \implies k = 21 a^6 b^4 \]

7. **Use the Condition \( a^3 + b^2 = 4 \):**
To maximize \( k \), we need to maximize \( a^6 b^4 \) under the constraint \( a^3 + b^2 = 4 \).

8. **Apply AM-GM Inequality:**
By the AM-GM inequality:
\[ \frac{a^3 + b^2}{2} \geq \sqrt{a^3 b^2} \]
Thus,
\[ 2 \geq \sqrt{a^3 b^2} \implies 4 \geq a^3 b^2 \implies (a^3 b^2)^{\frac{2}{5}} \leq \left(\frac{4}{2}\right)^{\frac{5}{2}} = 4 \]
Therefore,
\[ a^6 b^4 = (a^3 b^2)^{\frac{4}{5}} \leq 4^{\frac{4}{5}} = \frac{16}{4} = 16 \]

9. **Maximize \( k \):**
Thus, substituting back:
\[ k \leq 21 \cdot 16 = 336 \]

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