Let $$\alpha$$ be a root of the equation $$(a - c)x^2 + (b - a)x + (c - b) = 0$$, where $$a, b, c$$ are distinct real numbers such that the matrix $$\begin{pmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{pmatrix}$$ is singular. Then the value of $$\frac{(a - c)^2}{(b - a)(c - b)} + \frac{(b - a)^2}{(a - c)(c - b)} + \frac{(c - b)^2}{(a - c)(b - a)}$$ is

1. Analyze the Quadratic Equation

The given equation is:

$$(a - c)x^2 + (b - a)x + (c - b) = 0$$

Notice that the sum of the coefficients equals zero:

$$(a - c) + (b - a) + (c - b) = a - c + b - a + c - b = 0$$

Whenever the sum of the coefficients of a quadratic equation is zero, $$x = 1$$ is always a root. Thus, one of the roots is:

$$\alpha = 1$$

2. Verify the Matrix Singularity

If $$\alpha = 1$$, the matrix becomes:

$$\begin{pmatrix} 1^2 & 1 & 1 \\ 1 & 1 & 1 \\ a & b & c \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ a & b & c \end{pmatrix}$$

Since the first row and the second row are completely identical, the determinant of the matrix is zero. This confirms that the matrix is **singular** when $$\alpha = 1$$.

3. Simplify the Target Expression

To make the algebraic simplification easier, let us introduce three substitutions:

$$p = a - c$$

$$q = b - a$$

$$r = c - b$$

From our coefficient check in Step 1, we already know that:

$$p + q + r = 0$$

Now, substitute $$p, q,$$ and $$r$$ into the target expression:

$$\text{Value} = \frac{p^2}{qr} + \frac{q^2}{pr} + \frac{r^2}{pq}$$

Find a common denominator for the fractions:

$$\text{Value} = \frac{p^3 + q^3 + r^3}{pqr}$$

4. Apply the Algebraic Identity

Recall the conditional algebraic identity: if $$p + q + r = 0$$, then:

$$p^3 + q^3 + r^3 = 3pqr$$

Substitute this numerator back into our expression:

$$\text{Value} = \frac{3pqr}{pqr} = 3$$