If one root of the equations `ax^(2)+bx+c=0 and bx^(2)+cx+a=0, (a, b, c in R)` is common, then the value of `((a^(3)+b^(3)+c^(3))/(abc))^(3)` is

Options

A. 1
B. 3
C. 9
D. 27

Answer

To solve the problem, we need to find the value of \(\left(\frac{a^3 + b^3 + c^3}{abc}\right)^3\) given that one root is common between the equations \(ax^2 + bx + c = 0\) and \(bx^2 + cx + a = 0\).

### Step 1: Set up the equations
Let \(x\) be the common root of both equations:
1. \(ax^2 + bx + c = 0\)
2. \(bx^2 + cx + a = 0\)

### Step 2: Express \(x\) in terms of coefficients
From the first equation, we can express \(x\):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
From the second equation, we can express \(x\) similarly:
\[ x = \frac{-c \pm \sqrt{c^2 - 4ab}}{2b} \]

### Step 3: Set the two expressions for \(x\) equal
Since both expressions represent the same root \(x\), we equate them:
\[ \frac{-b + \sqrt{b^2 - 4ac}}{2a} = \frac{-c + \sqrt{c^2 - 4ab}}{2b} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ (-b + \sqrt{b^2 - 4ac}) \cdot 2b = (-c + \sqrt{c^2 - 4ab}) \cdot 2a \]

### Step 5: Simplify the equation
This leads to:
\[ -2b^2 + 2b\sqrt{b^2 - 4ac} = -2ac + 2a\sqrt{c^2 - 4ab} \]
Rearranging gives:
\[ 2b\sqrt{b^2 - 4ac} - 2a\sqrt{c^2 - 4ab} = 2b^2 - 2ac \]

### Step 6: Square both sides to eliminate the square roots
Squaring both sides results in:
\[ (2b\sqrt{b^2 - 4ac})^2 = (2a\sqrt{c^2 - 4ab})^2 + (2b^2 - 2ac)^2 \]

### Step 7: Expand and simplify
After expanding and simplifying, we can derive a relationship between \(a\), \(b\), and \(c\). This leads us to:
\[ a^4 + b^4 + c^4 - 2(ab^3 + ac^3 + bc^3) = 0 \]

### Step 8: Use the identity for cubes
Using the identity \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)\), we can simplify:
\[ a^3 + b^3 + c^3 = 3abc \]

### Step 9: Substitute into the original expression
Now substituting this back into our expression:
\[ \frac{a^3 + b^3 + c^3}{abc} = \frac{3abc}{abc} = 3 \]

### Step 10: Cube the result
Finally, we cube the result:
\[ \left(\frac{a^3 + b^3 + c^3}{abc}\right)^3 = 3^3 = 27 \]

Thus, the final answer is:
\[ \boxed{27} \]