question_answer 84)

. Using properties of determinant, prove that                   = 3(a + b + c) (ab + bc + ca)

Answer:

let       Operate             Take a + b + c as common from C1                   Operate  we get              Expand along C1        = (a + b + c) [(a + 2b) (a + 2c) ? (a ? c) (a ? b)]       = (a + b + c) [(a2 + 2ac + 2ab + 2ac) ? (a2 ? ac ?ab + bc)]       = (a + b + c) [a2 + 2ac + 2ab + 4bc ?  a2 + ab + ac ? bc]       = (a + b + c) [3ab + 3bc + 3ca]       = 3 (a + b + c) (ab + bc + ca)