Write down in terms of `x and n`, the term containing `x^3` in the expansion of `(1- (x)/(n) )^(n)` by the binomial theorem. if this term equals `(7)/(8)` when `x=-2, and n` is a positive integer, calculate the value of `n`.

Answer

To find the term containing \( x^3 \) in the expansion of \( \left(1 - \frac{x}{n}\right)^n \) using the Binomial Theorem, we can follow these steps:

### Step 1: Identify the general term
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, we have \( a = 1 \) and \( b = -\frac{x}{n} \). Thus, the general term \( T_k \) in the expansion is given by:
\[ T_k = \binom{n}{k} (1)^{n-k} \left(-\frac{x}{n}\right)^k = \binom{n}{k} \left(-\frac{x}{n}\right)^k \]

### Step 2: Find the term containing \( x^3 \)
We need to find the term where \( k = 3 \) because we want the term containing \( x^3 \). Therefore, we have:
\[ T_3 = \binom{n}{3} \left(-\frac{x}{n}\right)^3 = \binom{n}{3} \left(-\frac{x^3}{n^3}\right) \]

### Step 3: Substitute \( x = -2 \)
Now, we substitute \( x = -2 \) into the term:
\[ T_3 = \binom{n}{3} \left(-\frac{(-2)^3}{n^3}\right) = \binom{n}{3} \left(-\frac{-8}{n^3}\right) = \binom{n}{3} \frac{8}{n^3} \]

### Step 4: Set the term equal to \( \frac{7}{8} \)
According to the problem, this term equals \( \frac{7}{8} \):
\[ \binom{n}{3} \frac{8}{n^3} = \frac{7}{8} \]

### Step 5: Solve for \( n \)
First, we can rearrange the equation:
\[ \binom{n}{3} = \frac{7}{8} \cdot \frac{n^3}{8} \]
\[ \binom{n}{3} = \frac{7n^3}{64} \]
Using the formula for combinations:
\[ \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]
Setting both expressions equal gives:
\[ \frac{n(n-1)(n-2)}{6} = \frac{7n^3}{64} \]

### Step 6: Clear the fractions
Multiply both sides by \( 384 \) (the least common multiple of 6 and 64):
\[ 64n(n-1)(n-2) = 42n^3 \]

### Step 7: Simplify the equation
Expanding the left side:
\[ 64n(n^2 - 3n + 2) = 42n^3 \]
\[ 64n^3 - 192n^2 + 128n = 42n^3 \]
Rearranging gives:
\[ 22n^3 - 192n^2 + 128n = 0 \]
Factoring out \( n \):
\[ n(22n^2 - 192n + 128) = 0 \]

### Step 8: Solve the quadratic equation
We can ignore \( n = 0 \) since \( n \) is a positive integer. Now we solve:
\[ 22n^2 - 192n + 128 = 0 \]
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 22, b = -192, c = 128 \):
\[ n = \frac{192 \pm \sqrt{(-192)^2 - 4 \cdot 22 \cdot 128}}{2 \cdot 22} \]
Calculating the discriminant:
\[ n = \frac{192 \pm \sqrt{36864 - 11264}}{44} \]
\[ n = \frac{192 \pm \sqrt{25600}}{44} \]
\[ n = \frac{192 \pm 160}{44} \]

### Step 9: Calculate the two possible values for \( n \)
Calculating the two cases:
1. \( n = \frac{352}{44} = 8 \)
2. \( n = \frac{32}{44} = \frac{8}{11} \) (not valid since \( n \) must be an integer)

Thus, the only valid solution is:
\[ \boxed{8} \]