The coefficient of `x^5` in the expansion of `(2x^3-1/(3x)^2)^5` is

Options

A. `(26)/3`
B. `(80)/9`
C. `8`
D. `9`

Answer

To find the coefficient of \(x^5\) in the expansion of \(\left(2x^3 - \frac{1}{3x^2}\right)^5\), we can use the binomial theorem. The binomial expansion of \((a + b)^n\) is given by:

\[ \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In this case, let \(a = 2x^3\) and \(b = -\frac{1}{3x^2}\), and \(n = 5\).

### Step 1: Write the general term of the expansion
The general term \(T_r\) in the expansion is given by:

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

### Step 2: Simplify the general term
Now, we simplify \(T_r\):

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

This simplifies to:

\[ T_r = \binom{5}{r} 2^{5-r} (-1)^r \frac{1}{3^r} x^{3(5-r) - 2r} \]

### Step 3: Find the power of \(x\)
The power of \(x\) in \(T_r\) is:

\[ x^{15 - 3r - 2r} = x^{15 - 5r} \]

### Step 4: Set the power of \(x\) equal to 5
We need the power of \(x\) to be 5:

\[ 15 - 5r = 5 \]

### Step 5: Solve for \(r\)
Solving for \(r\):

\[ 15 - 5r = 5 \implies 10 = 5r \implies r = 2 \]

### Step 6: Substitute \(r\) back into the general term
Now, substitute \(r = 2\) back into the general term:

\[ T_2 = \binom{5}{2} 2^{5-2} (-1)^2 \frac{1}{3^2} x^{15 - 5 \cdot 2} \]

### Step 7: Calculate the coefficient
Calculating each part:

1. \(\binom{5}{2} = 10\)
2. \(2^{5-2} = 2^3 = 8\)
3. \((-1)^2 = 1\)
4. \(\frac{1}{3^2} = \frac{1}{9}\)
5. The power of \(x\) is \(x^5\).

Now, combine these:

\[ T_2 = 10 \cdot 8 \cdot 1 \cdot \frac{1}{9} x^5 = \frac{80}{9} x^5 \]

### Step 8: Conclusion
Thus, the coefficient of \(x^5\) in the expansion is:

\[ \frac{80}{9} \]