KeerthanaPosted on If a + b + c = 3 and none of a, b and c is equal to 1, what is the value of 1(1−a)(1−b)+1(1−b)(1−c)+1(1−c)(1−a)?\frac{1}{(1-a)(1-b)}+\frac{1}{(1-b)(1-c)}+\frac{1}{(1-c)(1-a)} ?(1−a)(1−b)1+(1−b)(1−c)1+(1−c)(1−a)1? a0 b1 c3 d6 Answer : Option AExplanation : a + b + c = 3 ∴1(1−a)(1−b)+1(1−b)(1−c)+1(1−c)(1−a)\therefore \frac{1}{(1-a)(1-b)}+\frac{1}{(1-b)(1-c)}+\frac{1}{(1-c)(1-a)}∴(1−a)(1−b)1+(1−b)(1−c)1+(1−c)(1−a)1 =1−c+1−a+1−b(1−a)(1−b)(1−c)=3−(a+b+c)(1−a)(1−b)(1−c)=\frac{1-c+1-a+1-b}{(1-a)(1-b)(1-c)}=\frac{3-(a+b+c)}{(1-a)(1-b)(1-c)}=(1−a)(1−b)(1−c)1−c+1−a+1−b=(1−a)(1−b)(1−c)3−(a+b+c) =3−3(1−a)(1−b)(1−c)=0=\frac{3-3}{(1-a)(1-b)(1-c)}=0=(1−a)(1−b)(1−c)3−3=0 Rate This:NaN / 5 - 1 votesAdd comment
