Assuming x to be so small that `x^(3)` and higher powers of x can be neglected, then value of
`E=(1-3/2x)^(5)(2+3x)^(6)`, is

Options

A. 64 + 96x + 720`x^(2)`
B. 65 - 97x + 721`x^(2)`
C. 64- 96x + 720`x^(2)`
D. 64 + 96x - 720`x^(2)`

Answer

To find the value of \( E = \left(1 - \frac{3}{2}x\right)^{5} \left(2 + 3x\right)^{6} \) while neglecting \( x^3 \) and higher powers, we can use the binomial theorem to expand both terms.

### Step-by-Step Solution:

1. **Expand \( \left(1 - \frac{3}{2}x\right)^{5} \)**:
Using the binomial theorem:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
Here, \( x = 1 \), \( y = -\frac{3}{2}x \), and \( n = 5 \).
\[ \left(1 - \frac{3}{2}x\right)^{5} = \binom{5}{0}(1)^{5}\left(-\frac{3}{2}x\right)^{0} + \binom{5}{1}(1)^{4}\left(-\frac{3}{2}x\right)^{1} + \binom{5}{2}(1)^{3}\left(-\frac{3}{2}x\right)^{2} \]
Neglecting higher powers:
\[ = 1 - \frac{15}{2}x + \frac{45}{4}x^2 \]

2. **Expand \( \left(2 + 3x\right)^{6} \)**:
Again using the binomial theorem:
\[ \left(2 + 3x\right)^{6} = \binom{6}{0}(2)^{6}(3x)^{0} + \binom{6}{1}(2)^{5}(3x)^{1} + \binom{6}{2}(2)^{4}(3x)^{2} \]
Neglecting higher powers:
\[ = 64 + 6 \cdot 32 \cdot 3x + 15 \cdot 16 \cdot 9x^2 \]
\[ = 64 + 576x + 2160x^2 \]

3. **Combine the expansions**:
Now, we multiply the two expansions:
\[ E = \left(1 - \frac{15}{2}x + \frac{45}{4}x^2\right) \left(64 + 576x + 2160x^2\right) \]
We will calculate the constant term, the coefficient of \( x \), and the coefficient of \( x^2 \).

- **Constant term**:
\[ 1 \cdot 64 = 64 \]

- **Coefficient of \( x \)**:
\[ -\frac{15}{2} \cdot 64 + 576 = -480 + 576 = 96 \]

- **Coefficient of \( x^2 \)**:
\[ \frac{45}{4} \cdot 64 + \left(-\frac{15}{2} \cdot 576\right) + 2160 \]
\[ = 720 - 4320 + 2160 = 720 - 2160 = -1440 \]

4. **Final expression**:
Combining all the terms, we get:
\[ E = 64 + 96x - 1440x^2 \]

### Final Result:
Thus, the value of \( E \) is:
\[ E = 64 + 96x - 1440x^2 \]