If a, b and c are real numbers, and `Delta=|b+cc+a a+b c+a a+bb+c a+bb+cc+a|=0`.Show that either `a" "+" "b" "+" "c" "=" "0" "or" "a" "=" "b" "=" "c`.

If a, b and c are real numbers, and `Delta=|b+cc+a a+b c+a a+bb+c a+bb

1.	If a, b and c are real numbers, and `Delta=\|b+cc+a a+b c+a a+bb+c a+bb+cc+a\|=0`.Show that either `a" "+" "b" "+" "c" "=" "0" "or" "a" "=" "b" "=" "c`.
Answer» ` Delta=\|{:(b+c,,c+a,,a+b),(c+a,,a+b,,b+c),(a+b,,b+c,,c+a):}\|` Applying `R_(1) to R_(1)+R_(2)+R_(3)` we have ` Delta=\|{:(2(a+b+c),,2(a+b+c),,2(a+b+c)),(c+a,,a+b,,b+c),(a+b,,b+c,,c+a):}\|` `=2 (a+b+c) \|{:(1,,1,,1),(c+a,,b-c,,b-a),(a+b,,c-a,,c-b):}\|` Applying `C_(2) to C_(2)-C_(1) " and "C_(3) to C_(3)-C_(1)` we have `Delta =2(a+b+c) \|{:(1,,0,,0),(c+a,,b-c,,b-a),(a+b,,c-a,,c-b),(,,,,):}\|` Expanding along `R_(1)` we have `Delta =2 (a+b+c) (1) [(b-c)(c-b)-(b-a)(c-a)]` `=2 (a+b+c) [-b^(2)-c^(2)+2bc-bc+ba+ac-a^(2)]` `=2 (a+b+c) [ab+bc+ca-a^(2)-b^(2)-c^(2)]` It is given that `Delta =0` Therefore, `(a+b+c) [ab+bc+ca-a^(2)-b^(2)-c^(2)]=0` `" or " (1//2)(a+b+c) [(a-b)^(2)+(b-c)^(2)+(c-a)^(2)=0` `" or Either " a+b+c =0` ` " or " (a-b)^(2) +(b-c)^(2) +(c-a)^(2)=0` ` rArr " Either " a+b+c =0 " or " a=b=c`