If ` x + 1/x = a, x^(2)+1/x^(3) = b," then "x^(3) + 1/x^(2)` is-

Options

A. `a^(3)+a^(2)-3a-2-b`
B. `a^(3)-a^(2)-3a+4-b`
C. `a^(3)-a^(2)+3a-6-b`
D. `a^(3)+a^(2)+3a-16-b`

Explanation

To solve the problem, we need to find the expression \( x^3 + \frac{1}{x^2} \) given that \( x + \frac{1}{x} = a \) and \( x^2 + \frac{1}{x^3} = b \).

### Step 1: Express \( x^2 + \frac{1}{x^2} \) in terms of \( a \)

We start with the equation:
\[ x + \frac{1}{x} = a \]
Squaring both sides gives us:
\[ \left( x + \frac{1}{x} \right)^2 = a^2 \]
Expanding the left side:
\[ x^2 + 2 + \frac{1}{x^2} = a^2 \]
Rearranging this, we find:
\[ x^2 + \frac{1}{x^2} = a^2 - 2 \]

### Step 2: Express \( x^3 + \frac{1}{x^3} \) in terms of \( a \)

Next, we use the identity for cubes:
\[ x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3\left( x + \frac{1}{x} \right) \]
Substituting \( a \) into this identity:
\[ x^3 + \frac{1}{x^3} = a^3 - 3a \]

### Step 3: Relate \( x^3 + \frac{1}{x^2} \) to \( b \)

We know from the problem statement:
\[ x^2 + \frac{1}{x^3} = b \]
We can express \( \frac{1}{x^3} \) in terms of \( x^3 + \frac{1}{x^3} \):
\[ \frac{1}{x^3} = \frac{1}{x^2} \cdot \frac{1}{x} \]
Thus, we can write:
\[ x^2 + \frac{1}{x^3} = b \implies x^2 + \frac{1}{x^3} = b \]

### Step 4: Find \( x^3 + \frac{1}{x^2} \)

Now we want to find \( x^3 + \frac{1}{x^2} \). We can express this as:
\[ x^3 + \frac{1}{x^2} = (x^3 + \frac{1}{x^3}) + (1 - \frac{1}{x^3} + \frac{1}{x^2}) \]
Using our earlier results:
\[ x^3 + \frac{1}{x^2} = (a^3 - 3a) + (b - x^2) \]

### Final Expression

Combining everything, we have:
\[ x^3 + \frac{1}{x^2} = a^3 + a^2 - 3a - b - 2 \]

### Conclusion

Thus, the final answer for \( x^3 + \frac{1}{x^2} \) is:
\[ x^3 + \frac{1}{x^2} = a^3 + a^2 - 3a - b - 2 \]