Find the interval of x, for which the expansion of `(8 – 3x)^(3/2)` in terms of power of x is valid.

Options

A. `xgt(4)/(3)`
B. `|x|gt(8)/(3)`
C. `xlt(3)/(8)`
D. `xlt(8 )/(3)`

Answer

To find the interval of \( x \) for which the expansion of \( (8 - 3x)^{3/2} \) in terms of powers of \( x \) is valid, we can follow these steps:

### Step 1: Identify the Binomial Expansion Condition
For the binomial expansion \( (1 + u)^n \) to be valid when \( n \) is a fraction, the condition is that \( |u| < 1 \).

### Step 2: Rewrite the Expression
We can rewrite \( (8 - 3x)^{3/2} \) in a suitable form. Factor out \( 8 \):
\[ (8 - 3x)^{3/2} = 8^{3/2} \left(1 - \frac{3x}{8}\right)^{3/2} \]

### Step 3: Apply the Condition
Now, we need to apply the condition \( \left| -\frac{3x}{8} \right| < 1 \):
\[ \left| \frac{3x}{8} \right| < 1 \]

### Step 4: Solve the Inequality
This inequality can be solved as follows:
\[ -\frac{3x}{8} < 1 \quad \text{and} \quad -\frac{3x}{8} > -1 \]

1. From \( -\frac{3x}{8} < 1 \):
\[ -3x < 8 \implies x > -\frac{8}{3} \]

2. From \( -\frac{3x}{8} > -1 \):
\[ -3x > -8 \implies x < \frac{8}{3} \]

### Step 5: Combine the Results
Combining both inequalities, we get:
\[ -\frac{8}{3} < x < \frac{8}{3} \]

### Final Answer
Thus, the interval of \( x \) for which the expansion is valid is:
\[ \boxed{\left(-\frac{8}{3}, \frac{8}{3}\right)} \]