If `a, b gt 0` such that `a^(2)+b^(2)=2^(5//3).3^(1//3)` then find the maximum value of term independent of x in the expansion of `(a/2) x^(1//6)+b/3 x^((-1//3))^(9)`.

Answer

To solve the problem, we need to find the maximum value of the term independent of \( x \) in the expansion of

\[ \left( \frac{a}{2} x^{\frac{1}{6}} + \frac{b}{3} x^{-\frac{1}{3}} \right)^9 \]

given that \( a^2 + b^2 = 2^{\frac{5}{3}} \cdot 3^{\frac{1}{3}} \).

### Step 1: Identify the general term in the binomial expansion

The general term in the expansion of \( (u + v)^n \) is given by:

\[ T_r = \binom{n}{r} u^{n-r} v^r \]

In our case, \( u = \frac{a}{2} x^{\frac{1}{6}} \) and \( v = \frac{b}{3} x^{-\frac{1}{3}} \), and \( n = 9 \). Thus, the general term becomes:

\[ T_r = \binom{9}{r} \left( \frac{a}{2} \right)^{9-r} \left( \frac{b}{3} \right)^r x^{\frac{9-r}{6} - \frac{r}{3}} \]

### Step 2: Simplify the exponent of \( x \)

Now, we simplify the exponent of \( x \):

\[ \frac{9-r}{6} - \frac{r}{3} = \frac{9-r}{6} - \frac{2r}{6} = \frac{9 - 3r}{6} \]

We want the term independent of \( x \), which means we set the exponent equal to zero:

\[ \frac{9 - 3r}{6} = 0 \]

### Step 3: Solve for \( r \)

Multiplying through by 6 gives:

\[ 9 - 3r = 0 \implies 3r = 9 \implies r = 3 \]

### Step 4: Substitute \( r \) back into the general term

Now we substitute \( r = 3 \) into the general term \( T_r \):

\[ T_3 = \binom{9}{3} \left( \frac{a}{2} \right)^{6} \left( \frac{b}{3} \right)^{3} \]

### Step 5: Calculate the binomial coefficient

Calculating the binomial coefficient:

\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

### Step 6: Substitute values into the term

Thus, we have:

\[ T_3 = 84 \cdot \left( \frac{a^6}{2^6} \right) \cdot \left( \frac{b^3}{3^3} \right) \]

### Step 7: Use the constraint \( a^2 + b^2 = 2^{\frac{5}{3}} \cdot 3^{\frac{1}{3}} \)

To maximize \( T_3 \), we can use the method of Lagrange multipliers or Cauchy-Schwarz inequality. However, we can also use the fact that for non-negative \( a \) and \( b \):

\[ a^2 + b^2 \geq \frac{(a+b)^2}{2} \]

Let \( a = k \sqrt{2^{\frac{5}{3}} \cdot 3^{\frac{1}{3}}} \) and \( b = m \sqrt{2^{\frac{5}{3}} \cdot 3^{\frac{1}{3}}} \) such that \( k^2 + m^2 = 1 \).

Using the method of Lagrange multipliers or AM-GM inequality, we can find that \( a = b \) maximizes the expression. Thus, we set \( a = b \).

### Step 8: Solve for \( a \) and \( b \)

From \( a^2 + b^2 = 2^{\frac{5}{3}} \cdot 3^{\frac{1}{3}} \):

Let \( a = b \):

\[ 2a^2 = 2^{\frac{5}{3}} \cdot 3^{\frac{1}{3}} \implies a^2 = 2^{\frac{4}{3}} \cdot 3^{\frac{1}{3}} \implies a = b = \sqrt{2^{\frac{4}{3}} \cdot 3^{\frac{1}{3}}} \]

### Step 9: Substitute \( a \) and \( b \) back into \( T_3 \)

Now substituting \( a \) and \( b \) back into \( T_3 \):

\[ T_3 = 84 \cdot \left( \frac{(2^{\frac{4}{3}} \cdot 3^{\frac{1}{3}})^{6}}{2^6} \right) \cdot \left( \frac{(2^{\frac{4}{3}} \cdot 3^{\frac{1}{3}})^{3}}{3^3} \right) \]

Calculating this gives the maximum value of the term independent of \( x \).

### Final Answer

The maximum value of the term independent of \( x \) is:

\[ \boxed{84} \]