Let α be a root of the equation (a - c)x² + (b - a)x + (c - b) = 0 where a, b, c are distinct real numbers such that the matrix [ α ², α, 1; 1, 1, 1; a, b, c ] is singular. Then, the value of (a - c)²/(b - a)(c - b) + (b - a)²/(a - c)(c - b) + (c - b)²/(a - c)(b - a) is

Solution

Maths sir · 4 steps

Name the three differences p, q, r. They add to zero, and then the cube identity finishes the sum.

1. Name the differences. Let p = a - c, q = b - a, r = c - b. Adding them, p + q + r = 0. Since a, b, c are distinct, none of p, q, r is zero.
2. One common denominator. The sum is p²/qr + q²/pr + r²/pq. Multiply each term up to the denominator pqr, and the numerators become p³, q³ and r³.

   p²/qr + q²/pr + r²/pq = (p³ + q³ + r³)/pqr = 3pqr/pqr

3. Use the identity. When p + q + r = 0, we have p³ + q³ + r³ = 3pqr. The singular-matrix condition only fixes α. The sum never uses α, so you do not need that condition here.
4. So. The value is 3pqr/pqr = 3, which is option (a).

The idea to keep. When three cyclic differences like a - c, b - a, c - b appear, they add to zero. Watch for p³ + q³ + r³ = 3pqr hiding behind the fractions.

Answer from the NTA final answer key. Solution by PrepWiser.