1 1 1 a 2 b 2 c 2 a 3 b 3 c 3 =

Question

1 1 1 a 2 b 2 c 2 a 3 b 3 c 3 =, a 2 b 2 a - b + b 2 c 2 b - c + c 2 a 2 c - a, a 2 b 3 - c 3 + b 2 c 3 - a 3 + c 2 a 3 - b 3, a 3 b 2 - c 2 + b 3 c 2 - a 2 + c 3 a 2 - b 2, a b a 3 - b 3 + b c b 3 - c 3 + c a c 3 - a 3

MARKS App · Accepted Answer

Let ∆ = 1 1 1 a 2 b 2 c 2 a 3 b 3 c 3 Apply C 2 ↔ C 2 - C 1 and C 3 ↔ C 3 - C 1 ∆ = 1 0 0 a 2 b 2 - a 2 c 2 - a 2 a 3 b 3 - a 3 c 3 - a 3 Expanding along R 1 , we get ∆ = b 2 - a 2 c 3 - a 3 - c 2 - a 2 b 3 - a 3 ⇒ ∆ = b 2 c 3 - a 3 + c 2 a 3 - b 3 + a 2 b 3 - c 3

Search any question & find its solution

Looking for more such questions to practice?