Let `r, s, t` be the roots of the equation `x ^(3) + ax ^(2) +bx+c=0,` such that `(rs )^(2) + (st)^(2) + (rt)^(2) =b^(2)-kac,` then k =

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( k \) given the equation \( x^3 + ax^2 + bx + c = 0 \) with roots \( r, s, t \) such that:

\[ (rs)^2 + (st)^2 + (rt)^2 = b^2 - kac \]

### Step 1: Use Vieta's Formulas
From Vieta's formulas, we know:
- The sum of the roots:
\[ r + s + t = -a \]
- The sum of the product of the roots taken two at a time:
\[ rs + rt + st = b \]
- The product of the roots:
\[ rst = -c \]

### Step 2: Expand \( (r + s + t)^2 \)
We can expand \( (r + s + t)^2 \):
\[ (r + s + t)^2 = r^2 + s^2 + t^2 + 2(rs + rt + st) \]
Substituting \( r + s + t = -a \) and \( rs + rt + st = b \):
\[ (-a)^2 = r^2 + s^2 + t^2 + 2b \]
This simplifies to:
\[ a^2 = r^2 + s^2 + t^2 + 2b \]
Thus,
\[ r^2 + s^2 + t^2 = a^2 - 2b \]

### Step 3: Relate \( r^2 + s^2 + t^2 \) to \( (rs)^2 + (st)^2 + (rt)^2 \)
We can express \( (rs)^2 + (st)^2 + (rt)^2 \) using the identity:
\[ (rs)^2 + (st)^2 + (rt)^2 = (rs + rt + st)^2 - 2rst(r + s + t) \]
Substituting the values we have:
\[ (rs)^2 + (st)^2 + (rt)^2 = b^2 - 2(-c)(-a) \]
This simplifies to:
\[ (rs)^2 + (st)^2 + (rt)^2 = b^2 - 2ac \]

### Step 4: Set the two expressions equal
From the problem statement, we have:
\[ (rs)^2 + (st)^2 + (rt)^2 = b^2 - kac \]
Equating the two expressions we derived:
\[ b^2 - 2ac = b^2 - kac \]

### Step 5: Solve for \( k \)
By simplifying the equation:
\[ -2ac = -kac \]
Dividing both sides by \( -ac \) (assuming \( ac \neq 0 \)):
\[ 2 = k \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{2} \]