If `ax^3+ bx - c` is divisible by `x^2+bx+c`, then 'a' is a root of the equation

Options

A. `cx^(2) - bx - 1 = 0`
B. `ax^(2) - bx - 1 = 0`
C. `bx^(2) - ax - 1 = 0`
D. none of these

Answer

To solve the problem, we need to determine under what conditions the expression \( ax^3 + bx - c \) is divisible by \( x^2 + bx + c \). We will perform polynomial long division and analyze the remainder.

### Step 1: Set up the division
We are dividing \( ax^3 + bx - c \) by \( x^2 + bx + c \).

### Step 2: Perform the first division
1. Divide the leading term of the dividend \( ax^3 \) by the leading term of the divisor \( x^2 \):
\[ \frac{ax^3}{x^2} = ax \]
2. Multiply \( ax \) by the entire divisor:
\[ ax(x^2 + bx + c) = ax^3 + abx^2 + acx \]
3. Subtract this from the original polynomial:
\[ (ax^3 + bx - c) - (ax^3 + abx^2 + acx) = -abx^2 + (b - ac)x - c \]

### Step 3: Perform the second division
Now we divide \( -abx^2 + (b - ac)x - c \) by \( x^2 + bx + c \).
1. Divide the leading term \( -abx^2 \) by \( x^2 \):
\[ \frac{-abx^2}{x^2} = -ab \]
2. Multiply \( -ab \) by the entire divisor:
\[ -ab(x^2 + bx + c) = -abx^2 - ab^2x - abc \]
3. Subtract this from the previous result:
\[ (-abx^2 + (b - ac)x - c) - (-abx^2 - ab^2x - abc) = (b - ac + ab^2)x + (abc - c) \]

### Step 4: Set the remainder to zero
Since \( ax^3 + bx - c \) is divisible by \( x^2 + bx + c \), the remainder must equal zero:
1. Set the coefficient of \( x \) to zero:
\[ b - ac + ab^2 = 0 \quad \text{(1)} \]
2. Set the constant term to zero:
\[ abc - c = 0 \quad \text{(2)} \]

### Step 5: Solve the equations
From equation (2):
\[ c(ab - 1) = 0 \]
This gives us two cases:
1. \( c = 0 \)
2. \( ab = 1 \)

Assuming \( c \neq 0 \), we have \( ab = 1 \).

Substituting \( ab = 1 \) into equation (1):
\[ b - ac + b^2 = 0 \]
Rearranging gives:
\[ b + b^2 = ac \quad \text{(3)} \]

### Step 6: Determine the relationship between \( a \), \( b \), and \( c \)
From \( ab = 1 \), we can express \( b \) in terms of \( a \):
\[ b = \frac{1}{a} \]
Substituting into equation (3):
\[ \frac{1}{a} + \left(\frac{1}{a}\right)^2 = ac \]
Multiplying through by \( a^2 \):
\[ a + 1 = a^2c \]
Rearranging gives:
\[ a^2c - a - 1 = 0 \]

### Conclusion
Thus, \( a \) is a root of the equation:
\[ a^2c - a - 1 = 0 \]