If the centroid of the triangle formed by the points A(a,b), B(b,c) and C(c,a) is at the origin, then what is the value of `(a^(2))/(bc)+(b^(2))/(ca)+(c^(2))/(ab)?`

Options

A. 0
B. `a+b+c`
C. 3
D. `a^(3)+b^(3)+c^(3)`

Answer

To solve the problem, we need to find the value of \(\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}\) given that the centroid of the triangle formed by the points \(A(a,b)\), \(B(b,c)\), and \(C(c,a)\) is at the origin.

### Step-by-Step Solution:

1. **Find the coordinates of the centroid**:
The coordinates of the centroid \(G\) of a triangle with vertices at \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) are given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
For our points:
- \(A(a, b)\)
- \(B(b, c)\)
- \(C(c, a)\)

The coordinates of the centroid are:
\[ G\left(\frac{a + b + c}{3}, \frac{b + c + a}{3}\right) \]

2. **Set the coordinates of the centroid to the origin**:
Since the centroid is at the origin, we have:
\[ \frac{a + b + c}{3} = 0 \quad \text{and} \quad \frac{b + c + a}{3} = 0 \]
This implies:
\[ a + b + c = 0 \]

3. **Use the identity for cubes**:
We know from algebra that if \(a + b + c = 0\), then:
\[ a^3 + b^3 + c^3 = 3abc \]

4. **Substitute into the expression**:
We need to evaluate:
\[ \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} \]
To combine these fractions, we take the common denominator \(abc\):
\[ \frac{a^3 + b^3 + c^3}{abc} \]
Using the identity we derived:
\[ \frac{3abc}{abc} = 3 \]

5. **Final answer**:
Thus, the value of \(\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}\) is:
\[ \boxed{3} \]