Prove that `|[1+a,1,1],[1,1+b,1],[1,1,1+c]|=(abc)(1/a+1/b+1/c+1)=(bc+ca+ab+abc)`

Prove that `|[1+a,1,1],[1,1+b,1],[1,1,1+c]|=(abc)(1/a+1/b+1/c+1)=(bc+c

1.	Prove that `\|[1+a,1,1],[1,1+b,1],[1,1,1+c]\|=(abc)(1/a+1/b+1/c+1)=(bc+ca+ab+abc)`
Answer» The given determinant `=\|{:(1+a, " "1, " "1),(" "1, 1+b, " "1),(" "1, " "1, 1+c):}\|` `=(abc)\|{:((1)/(a)+1, (1)/(a), (1)/(a)),((1)/(b), (1)/(b)+1, (1)/(b)),((1)/(c), (1)/(c), (1)/(c)+1):}\|` `["taking a, b, c common from"R_(1), R_(2) " and " R_(3) "respectively"]` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1)\|{:(1, 1, 1),((1)/(b), (1)/(b)+1, (1)/(b)),((1)/(c), (1)/(c), (1)/(c)+1):}\|` `["applying "R_(1) to R_(1) + R_(2) + R_(3) " and taking out" ((1)/(a) + (1)/(b) + (1)/(c) +1) "common from"R_(1)]` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1)\|{:(0, 0, 1),(0, 1, (1)/(b)),(-1, -1, (1)/(c)+1):}\|["applying"C_(1) to (C_(1) -C_(3))"and"C_(2) to (C_(2) -C_(3))]` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1)(1)\|{:(0, 1),(-1, -1):}\|["expanding by 1st row"]` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1)1` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1) = (bc+ca+ab+abc)` Hence, the result follows.