Given the function `f(x)=x^(3)-1`. Find `(f(b)-f(a))/(b-a) (b ne a) and f((a+b)/(2))`

Answer

To solve the problem step by step, we will first calculate \((f(b) - f(a)) / (b - a)\) and then find \(f\left(\frac{a+b}{2}\right)\).

### Step 1: Calculate \(f(b)\) and \(f(a)\)
Given the function:
\[ f(x) = x^3 - 1 \]
We need to find \(f(b)\) and \(f(a)\):
\[ f(b) = b^3 - 1 \]
\[ f(a) = a^3 - 1 \]

### Step 2: Calculate \(f(b) - f(a)\)
Now, we can calculate \(f(b) - f(a)\):
\[ f(b) - f(a) = (b^3 - 1) - (a^3 - 1) = b^3 - a^3 \]

### Step 3: Use the difference of cubes formula
We can use the difference of cubes formula:
\[ b^3 - a^3 = (b - a)(b^2 + ab + a^2) \]
Thus, we have:
\[ f(b) - f(a) = (b - a)(b^2 + ab + a^2) \]

### Step 4: Divide by \(b - a\)
Now, we divide \(f(b) - f(a)\) by \(b - a\):
\[ \frac{f(b) - f(a)}{b - a} = \frac{(b - a)(b^2 + ab + a^2)}{b - a} \]
Since \(b \neq a\), we can cancel \(b - a\):
\[ \frac{f(b) - f(a)}{b - a} = b^2 + ab + a^2 \]

### Step 5: Calculate \(f\left(\frac{a+b}{2}\right)\)
Next, we need to find \(f\left(\frac{a+b}{2}\right)\):
\[ f\left(\frac{a+b}{2}\right) = \left(\frac{a+b}{2}\right)^3 - 1 \]

### Step 6: Expand \(\left(\frac{a+b}{2}\right)^3\)
Using the binomial expansion:
\[ \left(\frac{a+b}{2}\right)^3 = \frac{(a+b)^3}{8} \]
Expanding \((a+b)^3\):
\[ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
Thus,
\[ \left(\frac{a+b}{2}\right)^3 = \frac{a^3 + 3a^2b + 3ab^2 + b^3}{8} \]

### Step 7: Substitute back to find \(f\left(\frac{a+b}{2}\right)\)
Now substituting back:
\[ f\left(\frac{a+b}{2}\right) = \frac{a^3 + 3a^2b + 3ab^2 + b^3}{8} - 1 \]
To express it in a single fraction:
\[ f\left(\frac{a+b}{2}\right) = \frac{a^3 + 3a^2b + 3ab^2 + b^3 - 8}{8} \]

### Final Answers
1. \(\frac{f(b) - f(a)}{b - a} = b^2 + ab + a^2\)
2. \(f\left(\frac{a+b}{2}\right) = \frac{a^3 + 3a^2b + 3ab^2 + b^3 - 8}{8}\)