Let a, b and c be distinct and none of them is equal to 1 . If the lines x+a y+a=0 , b x+y+b=0 and c x+c y+1=0 are concurrent, then the value of a a-1 + b b-1 + c c-1 is

Step-by-step solution

Given equations are aligned & x+a y+a=0 & b x+y+b=0 and c x+c y+1=0 aligned The above three lines are concurrent, so aligned & | array lll 1 & a & a b & 1 & b c & c & 1 array |=0 & 1(1-b c)-a(b-b c)+a(b c-c)=0 & 1-b c-a b+a b c+a b c-a c=0 & 1+2 a b c=a b+b c+a c Now, a a-1 + b b-1 + c c-1 & (a-1)(b-1)(c-1) = & a(b-1)(c-1)+b(a-1)(c-1)+c(a-1)(b-1) a c-a b-a c+b+a b c = & a b c-a b-a c+a+a b c-a b-b c+b c+c a b c+a+b+c-a b-b c-a c-1 = & 3 a b c-2(a b+b c+a c)+a+b+c a b c+a+b+c-(a b+b c+a c)-1 = & 3 a b c-2(1+2 a b c)