If a, b, c are real and `x^3-3b^2x+2c^3` is divisible by x -a and x - b, then (a) a =-b=-c (c) a = b = c or a =-2b-_ 2c (b) a = 2b = 2c (d) none of these

Options

A. a = - b = - c
B. a = 2b = 2c
C. a = b = c or a = - 2b = -2c
D. none of these

Answer

To solve the problem, we need to determine the conditions under which the polynomial \( f(x) = x^3 - 3b^2x + 2c^3 \) is divisible by both \( x - a \) and \( x - b \). This means that both \( f(a) = 0 \) and \( f(b) = 0 \).

### Step 1: Set up the equations
Since \( f(x) \) is divisible by \( x - a \) and \( x - b \), we can substitute \( x = a \) and \( x = b \) into the polynomial:

1. \( f(a) = a^3 - 3b^2a + 2c^3 = 0 \) (Equation 1)
2. \( f(b) = b^3 - 3b^2b + 2c^3 = 0 \) (Equation 2)

### Step 2: Simplify Equation 2
From Equation 2, we simplify:

\[ b^3 - 3b^3 + 2c^3 = 0 \implies -2b^3 + 2c^3 = 0 \implies c^3 = b^3 \]

This implies:

\[ c = b \quad \text{(since } a, b, c \text{ are real numbers)} \]

### Step 3: Substitute \( c \) back into Equation 1
Now that we know \( c = b \), we substitute \( c \) into Equation 1:

\[ a^3 - 3b^2a + 2b^3 = 0 \]

### Step 4: Factor the equation
We can rewrite \( 2b^3 \) as \( 2b \cdot b^2 \):

\[ a^3 - 3b^2a + 2b \cdot b^2 = 0 \]

### Step 5: Rearranging the equation
Rearranging gives us:

\[ a^3 - 3b^2a + 2b^3 = 0 \]

### Step 6: Factor by grouping
We can factor this equation. Notice that we can group terms:

\[ a^3 - 3b^2a + 2b^3 = 0 \]

This can be factored as:

\[ (a - b)(a^2 + ab + b^2 - 2b^2) = 0 \]

This gives us two cases:

1. \( a - b = 0 \) which implies \( a = b \)
2. \( a^2 + ab + b^2 - 2b^2 = 0 \) simplifies to \( a^2 + ab - b^2 = 0 \)

### Step 7: Solve the quadratic equation
Using the quadratic formula for \( a \):

\[ a = \frac{-b \pm \sqrt{b^2 + 4b^2}}{2} = \frac{-b \pm \sqrt{5b^2}}{2} = \frac{-b \pm b\sqrt{5}}{2} \]

This gives us two solutions for \( a \):

1. \( a = b \)
2. \( a = -2b \)

### Conclusion
Thus, we have two relationships:

1. \( a = b = c \)
2. \( a = -2b \) and \( c = b \)

### Final Answer
The correct option is (c) \( a = b = c \) or \( a = -2b \) and \( c = b \).

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