Find the coefficient of `x^15` in the expansion of `(2x^12 - (3)/(x^3) )^5`

Answer

To find the coefficient of \( x^{15} \) in the expansion of \( (2x^{12} - \frac{3}{x^3})^5 \), we can use the Binomial Theorem. The general 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, \( a = 2x^{12} \), \( b = -\frac{3}{x^3} \), and \( n = 5 \).

### Step 1: Write the general term
The general term \( T_{r+1} \) in the expansion is:

\[ T_{r+1} = \binom{5}{r} (2x^{12})^{5-r} \left(-\frac{3}{x^3}\right)^r \]

### Step 2: Simplify the general term
Now, we simplify this term:

\[ T_{r+1} = \binom{5}{r} (2^{5-r} (x^{12})^{5-r}) \left(-3^r (x^{-3})^r\right) \]

This simplifies to:

\[ T_{r+1} = \binom{5}{r} 2^{5-r} (-3)^r x^{12(5-r)} x^{-3r} \]

Combining the powers of \( x \):

\[ T_{r+1} = \binom{5}{r} 2^{5-r} (-3)^r x^{60 - 15r} \]

### Step 3: Set the exponent of \( x \) equal to 15
We need to find the value of \( r \) such that the exponent of \( x \) is 15:

\[ 60 - 15r = 15 \]

### Step 4: Solve for \( r \)
Rearranging gives:

\[ 60 - 15 = 15r \implies 45 = 15r \implies r = 3 \]

### Step 5: Substitute \( r \) back into the general term
Now, we substitute \( r = 3 \) back into the general term to find the coefficient:

\[ T_{4} = \binom{5}{3} 2^{5-3} (-3)^3 x^{60 - 15 \cdot 3} \]

Calculating each part:

\[ \binom{5}{3} = 10, \quad 2^{2} = 4, \quad (-3)^3 = -27 \]

Thus,

\[ T_{4} = 10 \cdot 4 \cdot (-27) \cdot x^{15} \]

### Step 6: Calculate the coefficient
Now, we calculate the coefficient:

\[ 10 \cdot 4 \cdot (-27) = 10 \cdot (-108) = -1080 \]

### Final Answer
Therefore, the coefficient of \( x^{15} \) in the expansion is:

\[ \boxed{-1080} \]