Find the coefficient of `x^3` in `( sqrt(x^5) + (3)/(sqrt(x^3)) )^6`

Answer

To find the coefficient of \( x^3 \) in the expression \( \left( \sqrt{x^5} + \frac{3}{\sqrt{x^3}} \right)^6 \), we can follow these steps:

### Step 1: Rewrite the expression
The expression can be rewritten as:
\[ \left( x^{5/2} + 3x^{-3/2} \right)^6 \]

### Step 2: Use the Binomial Theorem
According to the Binomial Theorem, the \( r+1 \)th term in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x^{5/2} \), \( b = 3x^{-3/2} \), and \( n = 6 \).

### Step 3: Write the general term
The general term \( T_{r+1} \) in our expansion is:
\[ T_{r+1} = \binom{6}{r} (x^{5/2})^{6-r} (3x^{-3/2})^r \]
This simplifies to:
\[ T_{r+1} = \binom{6}{r} (x^{5(6-r)/2}) (3^r x^{-3r/2}) \]
Combining the powers of \( x \):
\[ T_{r+1} = \binom{6}{r} 3^r x^{\frac{30 - 5r - 3r}{2}} = \binom{6}{r} 3^r x^{\frac{30 - 8r}{2}} \]

### Step 4: Set the exponent of \( x \) to 3
We need the exponent of \( x \) to equal 3:
\[ \frac{30 - 8r}{2} = 3 \]
Multiplying both sides by 2:
\[ 30 - 8r = 6 \]
Rearranging gives:
\[ 8r = 30 - 6 = 24 \quad \Rightarrow \quad r = 3 \]

### Step 5: Find the coefficient for \( r = 3 \)
Now substitute \( r = 3 \) back into the term:
\[ T_{4} = \binom{6}{3} 3^3 x^{\frac{30 - 8 \cdot 3}{2}} = \binom{6}{3} 3^3 x^{3} \]
Calculating \( \binom{6}{3} \):
\[ \binom{6}{3} = \frac{6!}{3! \cdot 3!} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20 \]
Now, calculate \( 3^3 \):
\[ 3^3 = 27 \]
Thus, the coefficient is:
\[ \text{Coefficient} = \binom{6}{3} \cdot 27 = 20 \cdot 27 = 540 \]

### Final Answer
The coefficient of \( x^3 \) in the expansion is \( \boxed{540} \).