In the expansion of `(sqrt(x^5)+3/(sqrt(x^3)))^6` coefficient of `x^3` is (i)`0` (ii)`120` (iii)`420` (iv)`540`

Options

A. 0
B. 120
C. 420
D. 540

Answer

To find the coefficient of \( x^3 \) in the expansion of \( \left( \sqrt{x^5} + \frac{3}{\sqrt{x^3}} \right)^6 \), we will use the binomial theorem.

### Step-by-Step Solution:

1. **Identify the terms in the binomial expansion**:
The expression can be rewritten as:
\[ \left( x^{5/2} + 3x^{-3/2} \right)^6 \]

2. **Apply the binomial theorem**:
The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = x^{5/2} \), \( b = 3x^{-3/2} \), and \( n = 6 \).

Thus,
\[ T_r = \binom{6}{r} (x^{5/2})^{6-r} (3x^{-3/2})^r \]

3. **Simplify the general term**:
\[ T_r = \binom{6}{r} (x^{5(6-r)/2}) (3^r)(x^{-3r/2}) \]
Combining the powers of \( x \):
\[ T_r = \binom{6}{r} 3^r x^{\frac{30 - 5r - 3r}{2}} = \binom{6}{r} 3^r x^{\frac{30 - 8r}{2}} \]

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

5. **Find the coefficient when \( r = 3 \)**:
Substitute \( r = 3 \) back into the general term:
\[ T_3 = \binom{6}{3} 3^3 x^{\frac{30 - 8 \cdot 3}{2}} = \binom{6}{3} 3^3 x^{\frac{30 - 24}{2}} = \binom{6}{3} 3^3 x^{3} \]

6. **Calculate \( \binom{6}{3} \) and \( 3^3 \)**:
\[ \binom{6}{3} = 20 \quad \text{and} \quad 3^3 = 27 \]
Therefore,
\[ T_3 = 20 \cdot 27 \cdot x^3 \]

7. **Calculate the coefficient**:
\[ \text{Coefficient of } x^3 = 20 \cdot 27 = 540 \]

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