If three distinct real number a,b and c satisfy `a^2(a+p)=b^2(b+p)=c^2(c+p)`, where `pepsilonR`, then value of `bc+ca+ab` is :

Options

A. `-p`
B. p
C. 0
D. `p^(2)//2`

Explanation

To solve the problem, we start with the given condition:

\[ a^2(a + p) = b^2(b + p) = c^2(c + p) \]

Let’s denote this common value as \( \lambda \). Therefore, we can write:

1. \( a^2(a + p) = \lambda \)
2. \( b^2(b + p) = \lambda \)
3. \( c^2(c + p) = \lambda \)

From these equations, we can express each of \( a^2(a + p) \), \( b^2(b + p) \), and \( c^2(c + p) \) as follows:

\[ a^3 + pa^2 - \lambda = 0 \]
\[ b^3 + pb^2 - \lambda = 0 \]
\[ c^3 + pc^2 - \lambda = 0 \]

This means that \( a \), \( b \), and \( c \) are the roots of the cubic equation:

\[ x^3 + px^2 - \lambda = 0 \]

For a cubic equation of the form:

\[ Ax^3 + Bx^2 + Cx + D = 0 \]

the sum of the products of the roots taken two at a time (i.e., \( ab + bc + ca \)) can be found using Vieta's formulas:

\[ ab + ac + bc = \frac{C}{A} \]

In our cubic equation, we have:

- \( A = 1 \)
- \( B = p \)
- \( C = 0 \) (since there is no \( x \) term)
- \( D = -\lambda \)

Thus, we can apply Vieta's formulas:

\[ ab + ac + bc = \frac{0}{1} = 0 \]

Therefore, the value of \( bc + ca + ab \) is:

\[ \boxed{0} \]